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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 15 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 15 Maths Textbook Solution.

Answers (1)

Answer: 2

Hint: Use indefinite formula then put the limit to get the required answer

Given:\int_{\pi }^{0}\frac{1}{1+\sin x}dx

Solution:\int_{\pi }^{0}\frac{1}{1+\sin x}dx

Rationalizing,\int_{0}^{\pi}\left(\frac{1}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}\right) d x=\int_{0}^{\pi}\left(\frac{1-\sin x}{(1+\sin x)(1-\sin x)}\right)

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

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

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

\left.\begin{array}{l} =[\tan x]_{0}^{\pi}-[\sec x]_{0}^{\pi} \\ =[\tan \pi-\tan 0]-[\sec \pi-\sec 0] \end{array}\right]\left[\begin{array}{l} \tan \pi=\tan 0=0 \\ \sec \pi=-1, \sec 0=1 \end{array}\right]

=\left [ 0-0 \right ]-\left [ -1-1 \right ]

=0-\left ( -2 \right )

=2

Posted by

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