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  • Need solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 36 textbook solution.

Need solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 36 textbook solution.

Answers (1)

Answer:-  \frac{\pi^{2}}{2 \sqrt{2}}

Hints:-  You must know the integration rules of trigonometric functions.

Given:-  \int_{0}^{\pi} \frac{x}{1+\sin ^{2} x}+\cos ^{7} x \cdot d x

Solution : \int_{0}^{\pi} \frac{x}{1+\sin ^{2} x}+\cos ^{7} x \cdot d x                                 ....(1)

Then 

\begin{gathered} I=\int_{0}^{\pi} \frac{\pi-x}{1-\sin ^{2}(\pi-x)}+\cos ^{7}(\pi-x) \cdot d x \\ \quad I=\int_{0}^{\pi} \frac{\pi-x}{1+\sin ^{2} x}-\cos ^{7} x \cdot d x \end{gathered}              ....(2)

Adding (1) +(2)

\begin{aligned} &\left.2 I=\int_{0}^{\pi} \frac{x}{1+\sin ^{2} x}+\cos ^{7} x+\frac{\pi-x}{1+\sin ^{2} x}-\cos ^{7} x\right) \cdot d x \\ &2 I=\pi \int_{0}^{\pi} \frac{1}{1+\sin ^{2} x} d x \end{aligned}

Dividing numerator and denominator by \cos^{2} x

\begin{aligned} &2 I=\pi \int_{0}^{\pi} \frac{\sec ^{2} x}{\sec ^{2} x+\tan ^{2} x} \cdot d x \\ &2 I=\pi \int_{0}^{\pi} \frac{\sec ^{2} x}{1+2 \tan ^{2} x} \cdot d x \\ &2 I=\pi \int_{0}^{\pi / 2} \frac{\sec ^{2} x}{1+2 \tan ^{2} x} \cdot d x \end{aligned}                                          \left[\because \int_{0}^{2 a} f(x) d x=\int_{0}^{2} \int_{0}^{a} f(x)\right]

Put \tan x = z

Then \sec^{2}x.dx=dz\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \text{when}x\rightarrow 0,z\rightarrow 0

                                                                    x=\frac{\pi}{2},z\rightarrow \infty

\begin{aligned} &2 I=2 \pi \int_{0}^{\infty} \frac{d z}{1+(\sqrt{2} z)^{2}} \\ &\left.2 I=2 \pi \times \frac{\tan ^{-1} \sqrt{2} z}{\sqrt{2}}\right]_{0}^{\infty} \end{aligned}

\begin{aligned} &I=\frac{\pi}{\sqrt{2}} \times\left(\tan ^{-1} \infty-\tan ^{-1} 0\right) \\ &I=\frac{\pi}{\sqrt{2}} \times\left(\frac{\pi}{2}-0\right) \\ &I=\frac{\pi^{2}}{2 \sqrt{2}} \end{aligned}

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