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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 27 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 27 Maths Textbook Solution.

Answers (1)

Answer: \frac{\pi }{2}-1

Hint:Use indefinite formula then put the limit to solve this integral

Given: \int_{0}^{\frac{\pi }{2}}x\cos xdx

Solution:

\int_{0}^{\frac{\pi }{2}}x\cos xdx

Integrating by parts  =>  Let x be the 1st part and cos x be the 2nd part

\begin{aligned} &=\left[x \int \cos x d x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left\{\frac{d(x)}{d x} \int \cos x d x\right\} d x \\ &=[x \sin x]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}(1 \cdot \sin x) d x \end{aligned} \quad\left[\int \cos x d x=\sin x\right]

=\left[\frac{\pi}{2} \sin \frac{\pi}{2}-0 \times \sin 0\right]-\int_{0}^{\frac{\pi}{2}} \sin x d x \quad\left[\sin \frac{\pi}{2}=1, \sin 0=0\right]

\begin{aligned} &=\left[\frac{\pi}{2} \cdot 1-0\right]-[-\cos x]_{0}^{\frac{\pi}{2}} \; \; \; \; \; \; \; \; \; \quad\left[\int \sin x d x=-\cos x\right] \\ &=\frac{\pi}{2}+[\cos x]_{0}^{\frac{\pi}{2}} \\ &=\frac{\pi}{2}+\left[\cos \frac{\pi}{2}-\cos 0\right] \end{aligned} \quad\left[\cos \frac{\pi}{2}=0, \cos 0=1\right]

=\frac{\pi }{2}+\left [ 0-1 \right ]

=\frac{\pi }{2}-1

 

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