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  • Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 17 maths textbook solution

Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 17 maths textbook solution

Answers (1)

Answer:

\frac{\pi }{4}

Given:

\int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x

Hint:

You must know derivation of sinx, cosx and tanx.


Explanation:  

Let

\begin{aligned} &I=\int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\tan \theta}{(1+\tan \theta)\left(1+\tan ^{2} \theta\right)} \cdot \sec ^{2} \theta d \theta \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\tan \theta}{(1+\tan \theta)\left(\sec ^{2} \theta\right)} \cdot \sec ^{2} \theta d \theta \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\tan \theta+1-1}{1+\tan \theta} d \theta \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} 1 d \theta-\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan \theta} d \theta \\\\ &=\int_{0}^{\frac{\pi}{2}} 1 d \theta-\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\frac{\sin \theta}{\cos \theta}} d \theta, \quad\left[\therefore \tan \theta=\frac{\sin \theta}{\cos \theta}\right] \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} 1 d \theta-\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\cos \theta+\sin \theta} d \theta \\\\\ &=[\theta]_{0}^{\frac{\pi}{2}}-I_{1}, \quad \text { where } I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\cos \theta+\sin \theta} d \theta \end{aligned}

\begin{aligned} &=\left(\frac{\pi}{2}-0\right)-I_{1} \\\\ &=\frac{\pi}{2}-I_{1} \ldots(i) \end{aligned}

Now,

\begin{aligned} &I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\cos \theta+\sin \theta} d \theta \ldots(i i) \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\cos \left(\frac{\pi}{2}-\theta\right)}{\cos \left(\frac{\pi}{2}-\theta\right)+\sin \left(\frac{\pi}{2}-\theta\right)} d \theta \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sin \theta}{\sin \theta+\cos \theta} d \theta, \quad\left[\therefore \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right] \\\\ &I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{\sin \theta}{\sin \theta+\cos \theta} d \theta \ldots(i i i) \end{aligned}

Adding (ii) and (iii)

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

\begin{aligned} &=\frac{1}{2}[\theta]_{0}^{\frac{\pi}{2}} \\\\ &=\frac{1}{2}\left[\frac{\pi}{2}-0\right] \\\\ &=\frac{\pi}{4} \end{aligned}

Put in (i)

\begin{aligned} &I=\frac{\pi}{2}-\frac{\pi}{4} \\\\ &I=\frac{2 \pi-\pi}{4} \\\\ &I=\frac{\pi}{4} \end{aligned}

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