Ex 7.10 class 12 maths ncert solutions | class 12 maths exercise 7.10 | class 12 maths ncert solutions chapter 7 exercise 7.10 | exercise 7.10 class 12 maths ncert solutions | integrals class 12 ncert solutions
If you’re studying Class 12 Maths, then understanding ex 7.10 class 12 maths NCERT solutions is very important. This exercise is part of the integrals class 12 NCERT solutions, which help students solve integration problems easily. The class 12 maths exercise 7.10 includes questions that build your concepts and improve your problem-solving skills. With the help of exercise 7.10 class 12 maths NCERT solutions, students can learn how to apply different methods of integration. These class 12 maths NCERT solutions chapter 7 exercise 7.10 are useful for board exams and help in gaining confidence in calculus topics.
integrals class 12 ncert solutions || exercise 7.10 class 12 maths ncert solutions || ex 7.10 class 12 maths ncert solutions || class 12 maths exercise 7.10 || class 12 maths ncert solutions chapter 7 exercise 7.10
Exercise 7.10
1. Evaluate the integrals using substitution.
\(\int_{0}^{1} \frac{x}{x^{2}+1} d x\)
\(\int_{0}^{1} \frac{x}{x^{2}+1} d x\)
Answer
Given: \( \int_{0}^{1} \frac{x}{x^{2}+1} d x \)
Let \( x^{2}+1=\mathrm{t} \)
\(\Rightarrow 2 x\mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow x\mathrm{d}x=\frac{ 1 }{ 2 }\mathrm{dt}\)
When \( x=0, \mathrm{t}=1 \) and when \( x=1, \mathrm{t}=2 \)
\(\Rightarrow \int_{0}^{1} \frac{x}{x^{2}+1} d x=\int_{1}^{2} \frac{d t}{2 t}\)
\(=\frac{1}{2} \int_{1}^{2} \frac{d t}{t}\)
\(=\frac{1}{2}[\log |t|]_{1}^{2}\)
\(=\frac{1}{2}[\log 2-\log 1]\)
\(=\frac{1}{2} \log 2\)
integrals class 12 ncert solutions || exercise 7.10 class 12 maths ncert solutions || ex 7.10 class 12 maths ncert solutions || class 12 maths exercise 7.10 || class 12 maths ncert solutions chapter 7 exercise 7.10
2. Evaluate the integrals using substitution.
\(\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi d \phi\)
\(\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi d \phi\)
Answer
Given: \( \int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi d \phi \)
Let \( \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi d \phi=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{4} \phi \cos \phi d \phi \)
\(=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi}\left(\cos ^{5} \phi\right)^{2} \cos \phi d \phi\)
\(=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi}\left(1-\sin ^{2} \phi\right)^{2} \cos \phi d \phi\)
Also, let \( \sin \phi=t \Rightarrow \cos \phi d \phi=d t \)
when, \( \phi=0, \mathrm{t}=0 \) and when \( \phi=\frac{\pi}{2}, t=1 \)
\(\text { so, } \mathrm{I}=\int_{0}^{1} \sqrt{\mathrm{t}}\left(1-t^{2}\right)^{2} d t\)
\(=\int_{0}^{1} t^{\frac{1}{2}}\left(1+t^{4}-2 t^{2}\right)^{2} d t\)
\(=\int_{0}^{1}\left(t^{\frac{1}{2}}+t^{\frac{9}{2}}-2 t^{\frac{5}{2}}\right) d t\)
\(=\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{11}{2}}}{\frac{11}{2}}+\frac{2 t^{\frac{7}{2}}}{\frac{7}{2}}\right]_{0}^{1}\)
\(=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}\)
\(=\frac{154+42-132}{231}=\frac{64}{231}\)
3. Evaluate the integrals using substitution.
\(\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)
\(\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)
Answer
Given: \( \int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x \)
Let \( x=\tan \theta \Rightarrow \mathrm{d}x=\sec ^{2} \theta \mathrm{d} \theta \)
When, \( x=0, \theta=0 \) and when \( x=1, \theta=\frac{ \pi }{ 4 } \)
Let \( \mathrm{I}=\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x \)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \sin ^{-1}\left(\frac{2 \tan \theta}{\tan ^{2} \theta+1}\right) \sec ^{2} \theta d \theta\)
\(=\int_{0}^{\frac{\pi}{4}} \sin ^{-1}(\sin 2 \theta) \sec ^{2} \theta d \theta\)
\(=\int_{0}^{\frac{\pi}{4}} 2 \theta \sec ^{2} \theta d \theta\)
\(=2 \int_{0}^{\frac{\pi}{4}} \theta \sec ^{2} \theta d \theta\)
By applying product rule as,
\(\int u \cdot v d x=\int u \cdot v d x-\int \frac{d u}{d x} \cdot\left\{\int v d x\right\} d x\)
\(\Rightarrow \mathrm{I}=2\left[\theta \sec ^{2} \theta d \theta-\int \frac{d}{d \theta} \theta \cdot\left\{\sec ^{2} \theta d \theta\right\} d \theta\right]_{0}^{\frac{\pi}{4}}\)
\(=2\left[\theta \tan \theta-\int 1 \cdot \tan \theta \mathrm{d} \theta\right]_{0}^{\frac{\pi}{4}}\)
\(=2\left[\frac{\pi}{4} \tan \frac{\pi}{4}-\log \left|\sec \frac{\pi}{4}\right|-0+\log |\sec 0|\right]\)
\(=2\left[\frac{\pi}{4}-\log (\sqrt{2})+\log 1\right]\)
\(=2\left[\frac{\pi}{4}-\frac{1}{2} \log (2)\right]\)
\(=\frac{\pi}{2}+\log (2)\)
4. Evaluate the integrals using substitution.
\(\int_{0}^{2} \sqrt{x+2}\left(\text { Put } x+2=\mathrm{t}^{2}\right)\)
\(\int_{0}^{2} \sqrt{x+2}\left(\text { Put } x+2=\mathrm{t}^{2}\right)\)
Answer
Given: \( \int_{0}^{2} \sqrt{x+2} d x \)
Let \( x+2=\mathrm{t}^{2} \Rightarrow \mathrm{d}x=2 \mathrm{tdt} \)
And \( x=\mathrm{t}^{2}-2 \)
when, \( x=0, t=\sqrt{2} \) and when \( x=2, t=2 \)
\( \int_{0}^{2} \sqrt{x+2} d x=\int_{\sqrt{2}}^{2}\left(t^{2}-2\right) \sqrt{t^{2}} 2 t d t\)
\(= 2 \int_{\sqrt{2}}^{2}\left(t^{2}-2\right) t \cdot t d t\)
\(= 2 \int_{\sqrt{2}}^{2}\left(t^{2}-2\right) t^{2} d t\)
\(= 2 \int_{\sqrt{2}}^{2}\left(t^{4}-2 t^{2}\right) d t\)
\(= 2\left[\frac{t^{5}}{5}-\frac{2 t^{3}}{3}\right]_{\sqrt{2}}^{2}\)
\(= 2\left[\frac{(2)^{5}}{5}-\frac{2(2)^{3}}{3}-\frac{\sqrt{(2)^{5}}}{5}+\frac{2 \sqrt{(2)^{3}}}{3}\right]_{\sqrt{2}}^{2}\)
\(= 2\left[\frac{32}{5}-\frac{16}{3}-\frac{4 \sqrt{2}}{5}+\frac{4 \sqrt{2}}{3}\right]\)
\(= 2\left[\frac{96-80-12 \sqrt{2}+20 \sqrt{2}}{15}\right]\)
\(= 2\left[\frac{16+8 \sqrt{2}}{15}\right]\)
\(= {\left[\frac{16(2+\sqrt{2})}{15}\right] }\)
\(= \frac{16 \sqrt{2}(\sqrt{2}+1)}{15}\)
integrals class 12 ncert solutions || exercise 7.10 class 12 maths ncert solutions || ex 7.10 class 12 maths ncert solutions || class 12 maths exercise 7.10 || class 12 maths ncert solutions chapter 7 exercise 7.10
5. Evaluate the integrals using substitution.
\(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x\)
\(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x\)
Answer
Given: \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x \)
Let \( \cos x=\mathrm{t} \)
\( \Rightarrow-\sin x \mathrm{d}x=\mathrm{dt} \)
\( \Rightarrow \sin x\mathrm{d}x=-\mathrm{dt} \)
When \( x=0, \mathrm{t}=1 \) and when \( x=\frac{ \pi }{ 2 }, \mathrm{t}=0 \)
\(=-\int_{1}^{0} \frac{d t}{1+t^{2}}\)
Because, \( \int \frac{d t}{x^{2}+a^{2}}=\frac{1}{a} \cdot \tan ^{-1} \frac{x}{a}+C \)
\( \Rightarrow-\int_{1}^{0} \frac{d t}{1+t^{2}}=-\left[\frac{1}{1} \cdot \tan ^{-1} t\right]_{1}^{0} \)
\( =-\left[\tan ^{-1} 0-\tan ^{-1} 1\right] \)
\( =-\left[0-\frac{\pi}{4}\right] \)
\( =-\left[-\frac{\pi}{4}\right] \)
\( =\frac{\pi}{4} \)
6. Evaluate the integrals using substitution.
\(\int_{0}^{2} \frac{d x}{x+4-x^{2}}\)
\(\int_{0}^{2} \frac{d x}{x+4-x^{2}}\)
Answer
Given: \( \int_{0}^{2} \frac{d x}{x+4-x^{2}} \)
\(\int_{0}^{2} \frac{d x}{x+4-x^{2}}=\int_{0}^{2} \frac{d x}{-\left(x^{2}-x-4\right)}\)
we can write it as, \( \int_{0}^{2} \frac{d x}{-\left(x^{2}-x-\frac{1}{4}-\frac{1}{4}-4\right)} \)
\(=\int_{0}^{2} \frac{d x}{-\left[\left(x-\frac{1}{2}\right)^{2}-\frac{17}{4}\right]}\)
\(=\int_{0}^{2} \frac{d x}{\left[\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}\right]}\)
Let \( x-\frac{1}{2}=t \Rightarrow d x=d t \)
When \( x=0, \mathrm{t}=-\frac{1}{2} \) and when \( x=2, \mathrm{t}=\frac{3}{2} \)
\(\Rightarrow \int_{0}^{2} \frac{d x}{\left[\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}\right]}=\int_{-\frac{1}{2}}^{\frac{3}{2}} \frac{d x}{\left[\left(\frac{\sqrt{17}}{2}\right)^{2}-(t)^{2}\right]}\)
\(\text { because, } \int \frac{d x}{\left[(a)^{2}-(x)^{2}\right]}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\)
\(\Rightarrow \int_{-\frac{1}{2}}^{\frac{3}{2}} \frac{d x}{\left.\left(\frac{\sqrt{17}}{2}\right)^{2}-(t)^{2}\right]}=\left[\frac{1}{2\left(\frac{\sqrt{17}}{2}\right)} \log \frac{\left(\frac{\sqrt{17}}{2}+t\right)}{\frac{\sqrt{17}}{2}-t}\right]_{-\frac{1}{2}}^{\frac{3}{2}}\)
\(=\frac{1}{\sqrt{17}}\left[\log \frac{\left(\frac{\sqrt{17}}{2}+\frac{3}{2}\right)}{\frac{\sqrt{17}}{2}-\frac{3}{2}}-\log \frac{\left(\frac{\sqrt{17}}{2}-\frac{1}{2}\right)}{\frac{\sqrt{17}}{2}+\frac{1}{2}}\right]\)
\(=\frac{1}{\sqrt{17}}\left[\log \frac{(\sqrt{17}+3)}{\sqrt{17}-3}-\log \frac{(\sqrt{17}-1)}{\sqrt{17}+1}\right]\)
\(=\frac{1}{\sqrt{17}}\left[\log \left\{\frac{(\sqrt{17}+3)}{\sqrt{17}-3} \times \frac{(\sqrt{17}-1)}{\sqrt{17}+1}\right\}\right]\)
\(=\frac{1}{\sqrt{17}}\left[\log \left\{\frac{(\sqrt{17}+3)(\sqrt{17}-1)}{(\sqrt{17}-3)(\sqrt{17}+1)}\right\}\right]\)
\(=\frac{1}{\sqrt{17}} \log \left[\frac{17+3+4 \sqrt{17}}{17+3-4 \sqrt{17}}\right]\)
\(=\frac{1}{\sqrt{17}} \log \left[\frac{20+4 \sqrt{17}}{20-4 \sqrt{17}}\right]\)
\(=\frac{1}{\sqrt{17}} \log \left[\frac{5+\sqrt{17}}{5-\sqrt{17}}\right]\)
\(=\frac{1}{\sqrt{17}} \log \left[\frac{(5+\sqrt{17})(5+\sqrt{17})}{(5-\sqrt{17})(5+\sqrt{17})}\right]\)
\(=\frac{1}{\sqrt{17}} \log \left[\frac{(25+17+10 \sqrt{17})}{25-\sqrt{17}}\right]\)
\(=\frac{1}{\sqrt{17}} \log \left[\frac{(42+10 \sqrt{17})}{8}\right]\)
\(=\frac{1}{\sqrt{17}} \log \left[\frac{(21+5 \sqrt{17})}{4}\right]\)
7. Evaluate the integrals using substitution.
\(\int_{-1}^{1} \frac{d x}{x^{2}+2 x+5}\)
\(\int_{-1}^{1} \frac{d x}{x^{2}+2 x+5}\)
Answer
Given: \( \int_{-1}^{1} \frac{d x}{x^{2}+2 x+5} \)
\(=\int_{-1}^{1} \frac{d x}{\left(x^{2}+2 x+1\right)+4}\)
\(=\int_{-1}^{1} \frac{d x}{(x+1)^{2}+(2)^{2}}\)
Let \( x+1=\mathrm{t} \)
\(\Rightarrow \mathrm{d}x=\mathrm{dt}\)
When \( x=-1, \mathrm{t}=0 \) and when \( x=1, \mathrm{t}=2 \)
\(\Rightarrow \int_{-1}^{1} \frac{d x}{(x+1)^{2}+(2)^{2}}=\int_{0}^{2} \frac{d t}{(t)^{2}+(2)^{2}}\)
because, \( \int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \cdot \tan ^{-1} \frac{x}{a}+C \)
\(\Rightarrow \int_{0}^{2} \frac{d t}{(t)^{2}+(2)^{2}}=\left[\frac{1}{2} \tan ^{-1} \frac{t}{2}\right]_{0}^{2}\)
\(=\frac{1}{2} \tan ^{-1} 1-\frac{1}{2} \tan ^{-1} 0\)
\(=\frac{1}{2}\left(\frac{\pi}{4}\right)=\frac{\pi}{8}\)
integrals class 12 ncert solutions || exercise 7.10 class 12 maths ncert solutions || ex 7.10 class 12 maths ncert solutions || class 12 maths exercise 7.10 || class 12 maths ncert solutions chapter 7 exercise 7.10
8. Evaluate the integrals using substitution.
\(\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x\)
\(\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x\)
Answer
Given: \( \int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x \)
Let \( 2 x=\mathrm{t} \Rightarrow 2 \mathrm{d}x=\mathrm{dt} \)
When \( x=1, \mathrm{t}=2 \) and when \( x=2, \mathrm{t}=4 \)
\(\Rightarrow \int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x=\int_{2}^{4}\left(\frac{1}{\left(\frac{t}{2}\right)}-\frac{1}{2\left(\frac{t}{2}\right)^{2}}\right) e^{t}\left(\frac{d t}{2}\right)\)
\(=\frac{1}{2} \int_{2}^{4}\left(\frac{2}{t}-\frac{2}{t^{2}}\right) e^{t} d t\)
\(=\int_{2}^{4} \frac{1}{2} \cdot(2)\left(\frac{1}{t}-\frac{1}{t^{2}}\right) e^{t} d t\)
\(=\int_{2}^{4}\left(\frac{1}{t}-\frac{1}{t^{2}}\right) e^{t} d t\)
now, let \( \frac{ 1 }{ \mathrm{t} }=\mathrm{f}(\mathrm{t}) \)
then, \( \mathrm{f}^{\prime}(\mathrm{t})=-\frac{ 1 }{ \mathrm{t}^{2} } \)
\(\Rightarrow \int_{2}^{4}\left(\frac{1}{t}-\frac{1}{t^{2}}\right) e^{t} d t=\int_{2}^{4}\left(\mathrm{f}(\mathrm{t})+\mathrm{f}^{\prime}(\mathrm{t})\right) \mathrm{e}^{\mathrm{t}} \mathrm{dt}\)
Because, \( \int_{2}^{4}\left(\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right) e^{2 x} d x=e^{x} f(x)+C \)
\(\Rightarrow \int_{2}^{4}\left(\mathrm{f}(\mathrm{t})+\mathrm{f}^{\prime}(\mathrm{t})\right) \mathrm{e}^{\mathrm{t}} \mathrm{dt}=\left[e^{t} \mathrm{f}(\mathrm{t})\right]_{2}^{4}\)
\(=\left[e^{t} \frac{1}{\mathrm{t}}\right]_{2}^{4}\)
\(=\frac{e^{4}}{4}-\frac{e^{2}}{2}\)
\(=\frac{e^{4}-2 e^{2}}{4}\)
\(=\frac{e^{4}\left(e^{2}-2\right)}{4}\)
9. The value of the integral \( \int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} \) is
A. 6 B. 0 C. 3 D. 4
A. 6 B. 0 C. 3 D. 4
Answer
Given: \( \int_{\frac{1}{3}}^{1}\left(\frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}}\right) d x \)
let \( \mathrm{I}=\int_{\frac{1}{3}}^{1}\left(\frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}}\right) d x \)
Now, let \( x=\sin \theta \Rightarrow \mathrm{d}x=\cos \theta \mathrm{d} \theta \)
when, \( x=\frac{1}{3}, \theta=\sin ^{-1}\left(\frac{1}{3}\right) \) and when \( x=1, \theta=\frac{ \pi }{ 2 } \)
\(\Rightarrow \mathrm{I}=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}\left(\frac{\left(\sin \theta-\sin ^{3} \theta\right)^{\frac{1}{3}}}{\sin ^{4} \theta}\right) \cos \theta \mathrm{d} \theta\)
\(=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}\left(\frac{(\sin \theta)^{\frac{1}{3}}\left(1-\sin ^{2} \theta\right)^{\frac{1}{3}}}{\sin ^{4} \theta}\right) \cos \theta \mathrm{d} \theta\)
\(=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}\left(\frac{(\sin \theta)^{\frac{1}{3}}\left(\cos ^{2} \theta\right)^{\frac{1}{3}}}{\sin ^{4} \theta}\right) \cos \theta \mathrm{d} \theta\)
\(=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}\left(\frac{(\sin \theta)^{\frac{1}{3}}\left(\cos ^{2} \theta\right)^{\frac{2}{3}}}{\sin ^{2} \theta \cdot \sin ^{2} \theta}\right) \cos \theta \mathrm{d} \theta\)
\(=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}\left(\frac{\left(\cos ^{2} \theta\right)^{\frac{2}{3}+1}}{(\sin \theta)^{2-\frac{1}{3}}}\right) \frac{1}{\sin ^{2} \theta} \mathrm{d} \theta\)
\(=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}\left(\frac{(\cos \theta)^{\frac{5}{3}}}{(\sin \theta)^{\frac{5}{3}}}\right) \cdot \operatorname{cosec}^{2} \theta \mathrm{d} \theta\)
\(=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}\left((\cot \theta)^{\frac{5}{3}}\right) \cdot \operatorname{cosec}^{2} \theta \mathrm{d} \theta\)
Now, let \( \cot \theta=\mathrm{t} \Rightarrow-\operatorname{cosec} 2 \theta \mathrm{d} \theta \)
when, \( \theta=\sin ^{-1}\left(\frac{1}{3}\right), t=2 \sqrt{2} \) and when \( \theta=\frac{\pi}{2}, \mathrm{t}=0 \)
\(=\int_{2 \sqrt{2}}^{0}-(t)^{\frac{5}{3}} \cdot d t\)
\(=-\left[\frac{(t)^{\frac{5}{3}+1}}{\frac{5}{3}+1}\right]_{2 \sqrt{2}}^{0}\)
\(=-\left[\frac{(t)^{\frac{8}{3}}}{\frac{8}{3}}\right]_{2 \sqrt{2}}^{0}\)
\(=-\frac{3}{2}\left[(0)^{\frac{8}{3}}-(2 \sqrt{2})^{\frac{8}{3}}\right]\)
\(=-\frac{3}{8}\left[-(\sqrt{8})^{\frac{8}{3}}\right]\)
\(=\frac{3}{8}\left[(8)^{\frac{4}{3}}\right]\)
\(=\frac{3}{8}[16]\)
\(=6\)
Correct option is: (A)
10. If \( \mathrm{f}(x)=\int_{0}^{x} \mathrm{t} \sin t \
dt\mathrm{~then~} \mathrm{f}^{\prime}(x) \) is
A. \( \cos x+x \sin x \) B. \( x \sin x \) C. \( x \cos x \) D. \( \sin x+x \cos x \)
A. \( \cos x+x \sin x \) B. \( x \sin x \) C. \( x \cos x \) D. \( \sin x+x \cos x \)
Answer
Given: \( \mathrm{f}(x)=\int_{0}^{x} \mathrm{t} \sin \mathrm{tdt} \)
Applying product rule,
\(\Rightarrow \int u \cdot v d x=u \cdot \int v d x-\int \frac{d u}{d x} \cdot\left\{\int v d x\right\} d x\)
\(\text { So, } \mathrm{f}(x)=\mathrm{t} \int_{0}^{x} \sin \mathrm{tdt} \int_{0}^{x}\left\{\left(\frac{d}{d t} t\right) \cdot \int \sin t d t\right\} d t\)
\(=[t(-\cos t)]_{0}^{x}-\int_{0}^{x}(-\cos t) d t\)
\(=[-t(\cos t)+\sin t]_{0}^{x}\)
\(=-x \cos x+\sin x-0\)
\(\Rightarrow \mathrm{f}(x)=-x \cos x+\sin x\)
\(\Rightarrow \mathrm{f}^{\prime}(x)=-[\{x(-\sin x)\}+\cos x]+\cos x\)
\(\Rightarrow \mathrm{f}^{\prime}(x)=-\left[x \cdot \frac{d}{d x} \cos x+\cos x \cdot \frac{d}{d x} x+\frac{d}{d x} \sin x\right]\)
\(=x \sin x-\cos x+\cos x\)
\(=x \sin x\)
Correct answer is B.
