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  • Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 37

Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 37

Answers (1)

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Answer:  \frac{\pi}{4}

Hint: To solve this equation we have to change cot into tan

Given:   \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot ^{7} x} d x

Solution:

\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot ^{7} x} d x=-\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\frac{1}{\cot ^{7} x}} d x              I=\int_{0}^{\frac{\pi}{2}} \frac{\tan ^{7} x}{1+\tan ^{7} x} d x  ………….. (1)

I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot \left(0+\frac{\pi}{2}-x\right)^{3}} d x

I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan ^{7} x} d x  …….. (2)

Adding equation (1) and (2)

2 I=\int_{0}^{\frac{\pi}{2}} \frac{\tan ^{7} x}{1+\tan ^{7} x} d x+\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan ^{7} x} d x

\begin{gathered} 2 I=\int_{0}^{\frac{\pi}{2}} d x \\ \end{gathered}

2 I=[x]_{0}^{\frac{\pi}{2}}

\begin{aligned} &2 I=\frac{\pi}{2}-0 \\ & \end{aligned}

I=\frac{\pi}{2 \times 2} \\

I=\frac{\pi}{4}

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