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  • Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 29 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 29 maths textbook solution.

Answers (1)

Answer:- \pi^{2}

Hints:-  You must know the integral rules of logarithmic functions.

Given:-  \int_{-\pi}^{\pi} \frac{2 x(1+\sin x)}{1+\cos ^{2} x} d x

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

                   I_{1}\; \; \; \; \; \; \; \; \; \; \; \; \; \; I_{2}

I_{1}=0        [being an odd function]

          \begin{aligned} &I_{2}=2 \int_{0}^{\pi} \frac{2 x \cdot \sin x}{1+\cos ^{2} x} \cdot d x \\ &I_{2}=4 \int_{0}^{\pi} \frac{x \cdot \sin x}{1+\cos ^{2} x} \cdot d x \end{aligned}

Where I_{3}=4 I_{2}+0=4 I

Let  I_{3}=\int_{0}^{\pi} \frac{x \cdot \sin x}{1+\cos ^{2} x} \cdot d x

                  \begin{aligned} &=\int_{0}^{\pi} \frac{(\pi-x) \cdot \sin (\pi-x)}{1+\cos ^{2}(\pi-x)} \cdot d x \\ &=\int_{0}^{\pi} \frac{(\pi-x) \cdot \sin x}{1+\cos ^{2} x} \cdot d x \\ &=\int_{0}^{\pi} \frac{\pi \cdot \sin x}{1+\cos ^{2} x} \cdot d x-\int_{0}^{\pi} \frac{x \cdot \sin x}{1+\cos ^{2} x} \cdot d x \end{aligned}

                 \begin{aligned} &=\pi \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} \cdot d x-I_{3} \\ 2 I_{3} &=\pi \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} \cdot d x \end{aligned}

Put \cos x = t\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x\rightarrow 0,t=1

      -\sin x \cdot d x=d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad x \rightarrow \pi, t=-1

       \begin{aligned} &2 I_{3}=-\pi \int_{1}^{-1} \frac{d t}{1+t^{2}} \\ &=\pi\left[\tan ^{-1} x\right]_{-1}^{1} \\ &2 I_{3}=\frac{\pi^{2}}{2} \end{aligned}

      \begin{aligned} &I_{3}=\frac{\pi^{2}}{4} \\ &\therefore I=\pi^{2} \end{aligned}

 

Posted by

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