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  • Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 49

Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 49

Answers (1)

best_answer

Answer:  -\frac{\pi}{2}

Hint: To solve this equation we use By Part

Given:  \int_{0}^{\pi} \cos 2 x \log \sin x d x

Solution:  \int f(x) g(x) d x\\

                        \downarrow                

                        By Parts

\begin{aligned} &f(x) g(x) d x=\int \frac{d(f x)}{d x} \int g(x) d x(d x) \\ & \end{aligned}

f(x)=\log \sin x \\

g(x)=\cos 2 x

\begin{aligned} &\Rightarrow\left[\log (\sin x) \int \cos 2 x d x-\int \frac{d(\log \sin x)}{d x} \int \cos 2 x d x d x\right]_{0}^{\pi} \\ & \end{aligned}

{\left[\log (\sin x) \frac{\sin 2 x}{2}-\int \frac{1}{\sin x} \cos x \cdot \frac{\sin 2 x}{x} d x\right]_{0}^{\pi}}

\begin{aligned} &\Rightarrow\left[\frac{1}{2} \sin 2 x \cdot \log \sin x-\frac{1}{2} \int 2 \cos ^{2} x d x\right]_{2}^{\pi} \\ & \end{aligned}

\Rightarrow\left[\frac{1}{2} \sin 2 x \cdot \log \sin x-\frac{1}{2} \int(\cos 2 x-1)\right]_{2}^{\pi} \\

\Rightarrow\left[\left[\frac{1}{2} \sin 2 x \cdot \log (\sin x)-\frac{1}{2}\left(\frac{\sin 2 x}{2}+x\right)\right]_{0}^{\pi}\right]

\begin{aligned} &I=\left[\frac{1}{2} \sin 2 x \cdot \log (\sin x)-\frac{1}{2} \frac{\sin 2 x}{2}-\frac{x}{2}\right]_{0}^{\pi} \\ & \end{aligned}

\frac{1}{2} \sin 2 \pi \cdot \log (\sin \pi)-\frac{1}{2} \frac{\sin 2 \pi}{2}-\frac{\pi}{2}-\frac{1}{2} \sin 0 \log \sin 0+\frac{1}{2} \sin 0+0 \\

=0-0-\frac{\pi}{2}-0+0+0 \\

=-\frac{\pi}{2}

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