Ex 7.11 Class 12 Maths Ncert Solutions

Given: \( \int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x \)
let, \( \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x\quad\ldots \text{(1)} \)
as, \( \left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\} \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \cos ^{2}\left(\frac{\pi}{2}-x\right) d x \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \sin ^{2}(x) d x \quad\ldots \text{(2)}\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}\left[\sin ^{2}(x)+\cos ^{2}(x)\right] d x\)
\(\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}[1] d x\)
\(\Rightarrow 2 \mathrm{I}=[x]_{0}^{\frac{\pi}{2}}\)
\(\Rightarrow 2 \mathrm{I}=\frac{\pi}{2}-0\)
\(\Rightarrow 2 \mathrm{I}=\frac{\pi}{2}\)
\(\Rightarrow \mathrm{I}=\frac{\pi}{4}\)

Given: \( \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \)
Let, \( \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \quad\ldots \text{(1)}\)
as, \( \left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\} \)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x\quad\ldots \text{(2)}\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)
\(\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}[1] d x\)
\(\Rightarrow 2 \mathrm{I}=[x]_{0}^{\frac{\pi}{2}}\)
\(\Rightarrow 2 \mathrm{I}=\frac{\pi}{2}-0\)
\(\Rightarrow 2 \mathrm{I}=\frac{\pi}{2}\)
\(\Rightarrow \mathrm{I}=\frac{\pi}{4}\)

Given: \( \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x \)
Let \( \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x\quad\ldots \text{(1)} \)
as, \( \left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\} \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)}{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)+\cos ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)} d x \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{\frac{3}{2}} x}{\cos ^{\frac{3}{2}} x+\sin ^{\frac{3}{2}} x} d x \quad\ldots \text{(2)}\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x\)
\(\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}[1] d x\)
\(\Rightarrow 2 \mathrm{I}=[x]_{0}^{\frac{\pi}{2}}\)
\(\Rightarrow 2 \mathrm{I}=\frac{\pi}{2}-0\)
\(\Rightarrow 2 \mathrm{I}=\frac{\pi}{2}\)
\(\Rightarrow \mathrm{I}=\frac{\pi}{4}\)

Given: \( \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x d x}{\sin ^{5} x+\cos ^{5} x} d x \)
Let \( \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x d x}{\sin ^{5} x+\cos ^{5} x} d x \quad\ldots \text{(1)}\)
as, \( \left\{\int_{0}^{\mathrm{a}} \mathrm{f}(x) \mathrm{d}x=\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{a}-x) \mathrm{d}x\right\} \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5}\left(\frac{\pi}{2}-x\right)}{\sin ^{5}\left(\frac{\pi}{2}-x\right)+\cos ^{5}\left(\frac{\pi}{2}-x\right)} d x \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{5} x}{\cos ^{5} x+\sin ^{5} x} d x \quad\ldots \text{(2)}\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{5} x+\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x\)
\(\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}[1] d x\)
\(\Rightarrow 2 \mathrm{I}=[x]_{0}^{\frac{\pi}{2}}\)
\(\Rightarrow 2 \mathrm{I}=\frac{\pi}{2}-0\)
\(\Rightarrow 2 \mathrm{I}=\frac{\pi}{2}\)
\(\Rightarrow \mathrm{I}=\frac{\pi}{4}\)

Given: \( \int_{-5}^{5}|x+2| d x \)
As we can see that \( (x+2) \leq 0 \) on \( [-5,-2] \) and \( (x+2) \geq 0 \) on \( [-2,5] \)
\(\text {as, }\left\{\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{d}x=\int_{\mathrm{a}}^{\mathrm{c}} \mathrm{f}(x) \mathrm{d}x+\int_{\mathrm{c}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{d}x\right\}\)
\(\Rightarrow \mathrm{I}=\int_{-5}^{-2}(x+2) d x+\int_{-5}^{5}|x+2| d x\)
\(\Rightarrow \mathrm{I}=-\left[\frac{x^{2}}{2}+2 x\right]_{-5}^{-2}+\left[\frac{x^{2}}{2}+2 x\right]_{-5}^{5}\)
\(\Rightarrow \mathrm{I}=\left[\frac{(-2)^{2}}{2}+2(-2)-\frac{(-5)^{2}}{2}-2(-5)\right]+\)\(\left[\frac{(5)^{2}}{2}+2(5)-\frac{(-2)^{2}}{2}-2(-2)\right]\)
\(\Rightarrow \mathrm{I}=-\left[2-4-\frac{25}{2}+10\right]+\left[\frac{25}{2}+10-2+4\right]\)
\(\Rightarrow \mathrm{I}=-2+4+\frac{25}{2}-10+\frac{25}{2}+10-2+4\)
\(\Rightarrow \mathrm{I}=29\)

Given: \( \int_{2}^{8}|x-5| d x \)
As we can see that \( (x-5) \leq 0 \) on \( [2,5] \) and \( (x+2) \geq 0 \) on \( [5,8] \) as,
\(\left\{\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{d}x=\int_{\mathrm{a}}^{\mathrm{c}} \mathrm{f}(x) \mathrm{d}x+\int_{\mathrm{c}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{d}x\right\}\)
\(\Rightarrow \mathrm{I}=\int_{2}^{5}-(x-5) d x+\int_{5}^{8}(x-5) d x\)
\(\Rightarrow \mathrm{I}=-\left[\frac{x^{2}}{2}-5 x\right]_{2}^{5}+\left[\frac{x^{2}}{2}-5 x\right]_{5}^{8}\)
\(\Rightarrow \mathrm{I}=\left[\frac{(5)^{2}}{2}-5(5)-\frac{(2)^{2}}{2}+5(2)\right]+\)\(\left[\frac{(8)^{2}}{2}+5(8)-\frac{(5)^{2}}{2}+5(5)\right]\)
\(\Rightarrow \mathrm{I}=-\left[\frac{25}{2}-25-2+10\right]+\left[\frac{64}{2}-40-\frac{25}{2}+25\right]\)
\(\Rightarrow \mathrm{I}=-\frac{25}{2}+17+32-15+\frac{25}{2}\)
\(\Rightarrow \mathrm{I}=9\)

Given: \( \int_{0}^{1} x(1-x)^{n} d x \)
Let, \( \mathrm{I}=\int_{0}^{1} z(1-x)^{n} d x \)
as, \( \left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\} \)
\(\Rightarrow \mathrm{I}=\int_{0}^{1}(1-x)(1-(1-x))^{n} d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{1}(1-x)(x)^{n} d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{1}(x)^{n}-(x)^{n+1} d x\)
\(\Rightarrow \mathrm{I}=\left[\frac{(x)^{n+1}}{n+1}-\frac{(x)^{n+2}}{n+2}\right]_{0}^{1}\)
\(\Rightarrow \mathrm{I}=\left[\frac{1}{n+1}-\frac{1}{n+2}\right]\)
\(\Rightarrow \mathrm{I}=\left[\frac{(n+2)-(n+1)}{(n+1)(n+2)}\right]\)
\(\Rightarrow \mathrm{I}=\left[\frac{1}{(n+1)(n+2)}\right]\)

Given: \( \int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x \)
Let \( \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x\quad\ldots \text{(1)} \)
as, \( \left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\} \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x \)
As, \( \left\{\tan (A-B)=\frac{\tan (A)-\tan (B)}{1+\tan (A) \tan (B)}\right\} \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log \left(1+\frac{\tan \left(\frac{\pi}{4}\right)-\tan (x)}{1+\tan \left(\frac{\pi}{4}\right) \tan (x)}\right) d x \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan (x)}{1+\tan (x)}\right) d x \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log \left[\frac{2}{1+\tan (x)}\right] d x \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log [2] d x-\int_{0}^{\frac{\pi}{4}} \log [1+\tan (x)] d x \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log [2] d x-\mathrm{I}\{ \)from \( (1)\} \)
\(\Rightarrow 2 \mathrm{I}=[x \log 2]_{0}^{\frac{\pi}{4}}\)
\(\Rightarrow 2 \mathrm{I}=\frac{\pi}{4} \log 2-0\)
\(\Rightarrow \mathrm{I}=\frac{\pi}{8} \log 2\)

Given: \( \int_{0}^{2} x \sqrt{2-x} d x \)
Let \( \mathrm{I}=\int_{0}^{2} x \sqrt{2-x} d x \quad\ldots \text{(1)}\)
as, \( \left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\} \)
\(\Rightarrow \mathrm{I}=\int_{0}^{2}(2-x) \sqrt{2-(2-x)} d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{2}(2-x) \sqrt{(x)} d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{2}\left(2 x^{\frac{1}{2}}-x^{\frac{3}{2}}\right) d x\)
\(\Rightarrow \mathrm{I}=\left[2\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)-\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right]_{0}^{2}\)
\(\Rightarrow \mathrm{I}=\left[\frac{4}{3}\left(x^{\frac{3}{2}}\right)-\frac{2}{5}\left(x^{\frac{5}{2}}\right)\right]_{0}^{2}\)
\(\Rightarrow \mathrm{I}=\left[\frac{4}{3}\left((2)^{\frac{3}{2}}\right)-\frac{2}{5}\left((2)^{\frac{5}{2}}\right)\right]_{0}^{2}\)
\(\Rightarrow \mathrm{I}=\frac{4}{3} \times 2 \sqrt{2}-\frac{2}{5} \times 4 \sqrt{2}\)
\(\Rightarrow \mathrm{I}=\frac{8 \sqrt{2}}{3}-\frac{8 \sqrt{2}}{5}\)
\(\Rightarrow \mathrm{I}=\frac{40 \sqrt{2}-24 \sqrt{2}}{15}\)
\(\Rightarrow \mathrm{I}=\frac{16 \sqrt{2}}{15}\)

Given: \( \int_{0}^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x \)
Let \( \mathrm{I}=\int_{0}^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}}\{2 \log \sin x-\log (2 \sin x \cos x)\} d x \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}}\{2 \log \sin x-\log (2)-\log (\sin x) \)\(-\log (\cos x)\} d x \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}}\{\log \sin x-\log 2-\log \cos x\} d x \quad\ldots \text{(1)}\)
as, \( \left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\} \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}}\left\{\log \sin \left(\frac{\pi}{2}-x\right)-\log 2-\log \cos\right.\)\(\left. \left(\frac{\pi}{2}-x\right)\right\} d x \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}}\{\log \cos x-\log 2-\log \sin x\} d x\quad\ldots \text{(2)} \)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}(-\log 2-\log 2) d x\)
\(2 \mathrm{I}=-2 \log 2 \int_{0}^{\frac{\pi}{2}}(1) d x\)
\(\Rightarrow 2 \mathrm{I}=-2 \log 2 \int_{0}^{\frac{\pi}{2}}[1] d x\)
\(\Rightarrow 2 \mathrm{I}=-2 \log 2[x]_{0}^{\frac{\pi}{2}}\)
\(\Rightarrow 2 \mathrm{I}=-2 \log 2\left[\frac{\pi}{2}-0\right]\)
\(\Rightarrow 2 \mathrm{I}=-2 \log 2\left(\frac{\pi}{2}\right)\)
\(\Rightarrow \mathrm{I}=\frac{\pi}{2}(-\log 2)\)
\(\Rightarrow \mathrm{I}=\frac{\pi}{2}\left(\log \frac{1}{2}\right)\)

Given: \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin ^{2} x\right) d x \)
Let \( \mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin ^{2} x\right) d x \)
As we can see \( f(x)=\sin ^{2} x \) and \( f(-x)=\sin ^{2}(-x)=(\sin (-x))^{2}=(-\sin x)^{2}= \) \( \sin ^{2} x \).
i.e. \( f(x)=f(-x) \)
so, \( \sin ^{2} x \) is an even function.
It is also known that if \( f(x) \) is an even function then,
\(\left\{\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x\right\}\)
\(\Rightarrow \mathrm{I}=2 \cdot \int_{0}^{\frac{\pi}{2}}\left(\sin ^{2} x\right) d x\)
\(\Rightarrow \mathrm{I}=2 \cdot \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2} d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}}(1-\cos 2 x) d x\)
\(\Rightarrow \mathrm{I}=\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}\)
\(\Rightarrow \mathrm{I}=\frac{\pi}{2}-\sin \pi-0+\sin 0\)
\(\Rightarrow \mathrm{I}=\frac{\pi}{2}\)

Given: \( \int_{0}^{\pi} \frac{x}{1+\sin x} d x \)
Let \( \mathrm{I}=\int_{0}^{\pi} \frac{x}{1+\sin x} d x \quad\ldots \text{(1)}\)
as, \( \left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\} \)
\(\Rightarrow \mathrm{I}=\int_{0}^{\pi} \frac{(\pi-x)}{1+\sin (\pi-x)} d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{\pi} \frac{(\pi-x)}{1+\sin x} d x\quad\ldots \text{(2)}\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{\pi} \frac{(\pi-x)+x}{1+\sin x} d x\)
\(2 \mathrm{I}=\int_{0}^{\pi} \frac{\pi}{1+\sin x} d x\)
\(2 \mathrm{I}=\pi \int_{0}^{\pi} \frac{(1-\sin x)}{(1+\sin x)(1-\sin x)} d x\)
\(2 \mathrm{I}=\pi \int_{0}^{\pi} \frac{(1-\sin x)}{\cos ^{2} x} d x\)
\(2 \mathrm{I}=\pi \int_{0}^{\pi}\left\{\frac{1}{\cos ^{2} x}-\frac{\sin x}{\cos ^{2} x}\right\} d x\)
\(2 \mathrm{I}=\pi \int_{0}^{\pi}\left\{\sec ^{2} x-\tan x \sec x\right\} d x\)
\(\Rightarrow 2 \mathrm{I}=\pi[\tan x-\sec x]_{0}^{\pi}\)
\(\Rightarrow 2 \mathrm{I}=\pi[2]\)

Given: \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin ^{7} x\right) d x \)
Let \( \mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin ^{7} x\right) d x \)
As we can see \( f(x)=\sin ^{7} x \) and \( f(-x)=\sin ^{7}(-x)=(\sin (-x))^{2}=(-\sin x)^{7}= \) \( \sin ^{7} \mathrm{x} \).
i.e. \( f(x)=f(-x) \)
so, \( \sin ^{7} x \) is an even function.
It is also known that if \( f(x) \) is an even function then,
\(\left\{\int_{-\mathrm{a}}^{\mathrm{a}} \mathrm{f}(x) \mathrm{d}x=0\right\}\)
\(\Rightarrow \mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin ^{7} x\right) d x=0\)

Given: \( \int_{0}^{2 \pi}\left(\cos ^{2} x\right) d x \)
Let \( \mathrm{I}=\int_{0}^{2 \pi}\left(\cos ^{2} x\right) d x \)
As we see, \( f(x)=\cos ^{5} x \) and
\( f(2 \pi-x)=\cos ^{5}(2 \pi-x)=\cos ^{5} x=f(x) \)
Because, \( \int_{0}^{2 a} \mathrm{f}(x) \mathrm{d}x=\int_{0}^{a} \mathrm{f}(x) d x \),
if \( \mathrm{f}(2 \mathrm{a}-x)=\mathrm{f}(x) \)
And \( \int_{0}^{2 a} \mathrm{f}(x) \mathrm{d}x=0 \), if \( \mathrm{f}(2 \mathrm{a}-x)=-\mathrm{f}(x) \)
\(\Rightarrow \mathrm{I}=2 \cdot \int_{0}^{\pi}\left(\cos ^{2} x\right) d x\)
Now \( \left\{\cos ^{5}(\pi-x)=-\cos ^{5} x\right\} \)
\(\Rightarrow \mathrm{I}=0\)

Given: \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x \)
Let \( \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x\quad\ldots \text{(1)} \)
as, \( \left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\} \)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\cos x \sin x} d x\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x+\cos x-\sin x}{1+\sin x \cos x} d x\)
\(\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{0}{1+\sin x \cos x} d x\)
\(\Rightarrow \mathrm{I}=0\)

Given: \( \int_{0}^{\pi} \log (1+\cos x) d x \)
Let \( \mathrm{I}=\int_{0}^{\pi} \log (1+\cos x) d x\quad\ldots \text{(1)} \)
as, \( \left\{\int_{0}^{\mathrm{a}} \mathrm{f}(x) \mathrm{d}x=\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{a}-x) \mathrm{d}x\right\} \)
\(\Rightarrow \mathrm{I}=\int_{0}^{\pi} \log (1+\cos (\pi-x)) d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{\pi} \log (1-\cos x) d x\quad\ldots \text{(2)}\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{\pi}\{\log (1+\cos x)+\log (1-\cos x)\} d x\)
\(2 \mathrm{I}=\int_{0}^{\pi} \log \left(1-\cos ^{2} x\right) d x\)
\(2 \mathrm{I}=\int_{0}^{\pi} \log \left(\sin ^{2} x\right) d x\)
\(2 \mathrm{I}=\int_{0}^{\pi} 2 \cdot \log (\sin x) d x\)
\(2 \mathrm{I}=2 \cdot \int_{0}^{\pi} \log (\sin x) d x\)
\(I=\int_{0}^{\pi} \log (\sin x) d x\quad\ldots \text{(3)}\)
Because, \( \int_{0}^{2 a} \mathrm{f}(x) \mathrm{d}x=\int_{0}^{a} \mathrm{f}(x) d x \),
if \( \mathrm{f}(2 \mathrm{a}-x)=\mathrm{f}(x) \)
Here, if \( f(x)=\log (\sin x) \)
and \( f(\pi-x)=\log (\sin (\pi-x))=\log (\sin x)= \) \( \mathrm{f}(x) \)
\(\Rightarrow \mathrm{I}=2 \cdot \int_{0}^{\frac{\pi}{2}} \log (\sin x) d x\quad\ldots \text{(4)}\)
\(\Rightarrow \mathrm{I}=2 \cdot \int_{0}^{\frac{\pi}{2}} \log \sin \left(\frac{\pi}{2}-x\right) d x\)
\(\Rightarrow \mathrm{I}=2 \cdot \int_{0}^{\frac{\pi}{2}} \log (\cos x) d x\quad\ldots \text{(5)}\)
Adding (1) and (2), we get
\(\Rightarrow 2 \mathrm{I}=2 \cdot \int_{0}^{\frac{\pi}{2}}(\log \sin x+\log \cos x) d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}}(\log \sin x+\log \cos x+\log 2-\log 2) d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}}(\log (2 \sin x \cos x)-\log 2) d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log (\sin 2 x) d x-\int_{0}^{\frac{\pi}{2}} \log 2 d x\)
Let \( 2 x=\mathrm{t} \Rightarrow 2 \mathrm{d}x=\mathrm{dt} \)
When \( x=0, \mathrm{t}=0 \) and when \( x=\frac{\pi }{ 2}, \mathrm{t}=\pi \)
\(\Rightarrow \mathrm{I}=\left[\frac{1}{2} \int_{0}^{\pi}(\log (\sin t) d t)\right]-\left(\frac{\pi}{2} \log 2\right)\)
\(\Rightarrow \mathrm{I}=\left[\frac{1}{2}\right]-\left(\frac{\pi}{2} \log 2\right)\)
\(\Rightarrow \mathrm{I}=-\left(\frac{\pi}{2} \log 2\right)\)
\(\Rightarrow \mathrm{I}=-(\pi \log 2)\)

Given: \( \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a}-x} d x \)
Let \( \mathrm{I}=\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a}-x} d x\quad\ldots \text{(1)} \)
as, \( \left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\} \)
\(\Rightarrow \mathrm{I}=\int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x\quad\ldots \text{(2)}\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{a} \frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}} d x\)
\(\Rightarrow 2 \mathrm{I}=\int_{0}^{a}[1] d x\)
\(\Rightarrow 2 \mathrm{I}=[x]_{0}^{a}\)
\(\Rightarrow 2 \mathrm{I}=\mathrm{a}\)
\(\Rightarrow \mathrm{I}=\frac{a}{2}\)

Given: \( \int_{0}^{4}|x-1| d x \)
As we can see that \( (x-1) \leq 0 \) when \( 0 \leq x \leq 1 \) and \( (x-1) \geq 0 \) when \( 1 \leq x \leq 4 \)
\(\text {as, }\left\{\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{d}x=\int_{\mathrm{a}}^{\mathrm{c}} \mathrm{f}(x) \mathrm{d}x+\int_{\mathrm{c}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{d}x\right\}\)
\(\Rightarrow \mathrm{I}=\int_{0}^{1}-(x-1) d x+\int_{1}^{4}(x-1) d x\)
\(\Rightarrow \mathrm{I}=-\left[\frac{x^{2}}{2}-x\right]_{0}^{1}+\left[\frac{x^{2}}{2}-x\right]_{1}^{4}\)
\(\Rightarrow \mathrm{I}=-\left[\frac{(1)^{2}}{2}-1-\frac{(0)^{2}}{2}+0\right]+\left[\frac{(4)^{2}}{2}-4-\frac{(1)^{2}}{2}+1\right]\)
\(\Rightarrow \mathrm{I}=-\left[\frac{1}{2}-1\right]+\left[8-4-\frac{1}{2}+1\right]\)
\(\Rightarrow \mathrm{I}=\frac{1}{2}+5-\frac{1}{2}\)
\(\Rightarrow \mathrm{I}=5\)

Given: \( \int_{0}^{a} f(x) g(x) d x \)
Let \( \mathrm{I}=\int_{0}^{a} \mathrm{f}(x) g(x) \mathrm{d} x\quad\ldots \text{(1)}\)
as, \( \left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\} \)
\(\Rightarrow \mathrm{I}=\int_{0}^{a} \mathrm{f}(\mathrm{a}-x) \mathrm{g}(\mathrm{a}-x) \mathrm{d}x\quad\ldots \text{(2)}\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{a}\{\mathrm{f}(x) \mathrm{g}(x)+\mathrm{f}(x) \mathrm{g}(\mathrm{a}-x)\} \mathrm{d}x\)
\(\Rightarrow 2 \mathrm{I}=\int_{0}^{a} \mathrm{f}(x)\{\mathrm{g}(x)+\mathrm{g}(\mathrm{a}-x)\} \mathrm{d}x\)
\(\Rightarrow 2 \mathrm{I}=\int_{0}^{a} \mathrm{f}(x)\{4\} d x \text { as, }\{\mathrm{g}(x)=\mathrm{g}(\mathrm{a}-x)=4\}\)
\(\Rightarrow \mathrm{I}=\frac{1}{2} \int_{0}^{a} \mathrm{f}(x) \times 4 d x\)
\(\Rightarrow \mathrm{I}=2 \cdot \int_{0}^{a} \mathrm{f}(x) d x\)

Given: \( \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^{5} x+1\right) d x \)
Let, \( \mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^{5} x+1\right) d x \)
\(\Rightarrow \mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}\right) d x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x \cos x) d x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\)\(\left(\tan ^{5} x\right) d x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1) d x\)
It is also known that if \( f(x) \) is an even function then,
\(\left\{\int_{-\mathrm{a}}^{\mathrm{a}} \mathrm{f}(x) \mathrm{d}x=2 \int_{-\mathrm{a}}^{\mathrm{a}} \mathrm{f}(x) \mathrm{d}x\right\}\)
It is also known that if \( f(x) \) is an odd function then,
\(\Rightarrow \mathrm{I}=0+0+0+2 \cdot \int_{0}^{\frac{\pi}{2}}(1) d x\left\{\int_{-\mathrm{a}}^{\mathrm{a}} \mathrm{f}(x) \mathrm{d}x=0\right\}\)
\(\Rightarrow \mathrm{I}=2 \cdot[x]_{0}^{\frac{\pi}{2}}\)
\(\Rightarrow \mathrm{I}=2 \cdot \frac{\pi}{2}\)
\(\Rightarrow \mathrm{I}=\pi\)

Given: \( \int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x \)
Let, \( \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x \quad\ldots \text{(1)}\)
as, \( \left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\} \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left(\frac{\pi}{2}-x\right)}\right) d x \)
\( \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \cos x}{4+3 \sin }\right) d x \quad\ldots \text{(2)}\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}\left\{\log \left(\frac{4+3 \sin x}{4+3 \cos x}\right)+\left(\frac{4+3 \cos x}{4+3 \sin }\right)\right\} d x\)
\(\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}\left(\log \frac{4+3 \sin x}{4+3 \cos x} \times \frac{4+3 \cos x}{4+3 \sin }\right) d x\)
\(\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log 1 d x\)
\(\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} 0 \cdot d x\)
\(\Rightarrow \mathrm{I}=0\)