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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 22 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 22 Maths Textbook Solution.

Answers (1)

Answer:

\sqrt{2}\left[\tan ^{-1}\left(\frac{(\sqrt{2 \tan x}-1)}{\sqrt{2}}\right)+\tan ^{-1}\left(\frac{\sqrt{2 \tan x}+1}{\sqrt{2}}\right)\right]

Given:

\int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x

Hint:

To solve this equation we use partial function method.

Solution: 

\int \frac{\sin x+\cos x}{\sqrt{2 \sin x \cos x}} d x \quad \because\left[\sin x=\frac{\tan x}{\sec x}, \cos x=\frac{1}{\sec x}\right]

   \int \frac{\frac{\tan x}{\sec x}+\frac{1}{\sec x}}{\sqrt{\frac{\operatorname{stan} x}{\sec x} \times \frac{1}{\sec x}}} d x

\int \frac{\tan x+1}{\sqrt{2(\tan x)}} d x

   \frac{1}{\sqrt{2}} \int \frac{\tan x+1}{\sqrt{\tan x}} \frac{\tan ^{2} x+1}{\tan ^{2} x+1} d x

\frac{1}{\sqrt{2}} \int \frac{\sec ^{2} x(\tan x+1)}{\sqrt{\tan x}\left(\tan ^{2} x+1\right)} d x \quad\left[\because \tan x=u, \sec ^{2} x d x=d u\right]

=\frac{1}{\sqrt{2}} \int \frac{u+1}{\sqrt{u}\left(u^{2}+1\right)} d u

\left.=\frac{1}{\sqrt{2}} \int \frac{v^{2}+1}{\sqrt{u}\left(v^{4}+1\right)} 2 \sqrt{u} d v_{\cdots . .[\text { put } u}=v^{2}, d u=2 v d v\right]

=\sqrt{2} \int \frac{v^{2}+1}{v^{4}+1} d v

=\sqrt{2} \int \frac{v^{2}+1}{\left(v^{2}-\sqrt{2} v+1\right)\left(v^{2}+\sqrt{2} v+1\right)} d v

=\frac{v^{2}+1}{\left(v^{2}-\sqrt{2} v+1\right)\left(v^{2}+\sqrt{2} v+1\right)}=\frac{A}{v^{2}-\sqrt{2} v+1}+\frac{B}{v^{2}+\sqrt{2} v+1}

=\sqrt{2}\left(\frac{1}{2}\right)\left(\int \frac{1}{v^{2}-\sqrt{2} v+1} d v+\int \frac{1}{v^{2}+\sqrt{2} v+1} d v\right)

=\frac{\sqrt{2}}{\sqrt{2} \times \sqrt{2}} \int \frac{1}{v-\left(\frac{1}{\sqrt{2}}\right)^{2}+\frac{1}{2}} d v+\int \frac{1}{v+\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{2}\right)} d v

\left[\frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)\right]

=\frac{1}{\sqrt{2}} \sqrt{2} \frac{\tan ^{-1}\left(v-\left(\frac{1}{\sqrt{2}}\right)\right)}{\frac{1}{\sqrt{2}}}+\sqrt{2} \frac{\tan ^{-1}\left(v+\left(\frac{1}{\sqrt{2}}\right)\right)}{\frac{1}{\sqrt{2}}}

=\sqrt{2}\left[\tan ^{-1}\left(\frac{\sqrt{2 v}-1}{\sqrt{2}}\right)+\tan ^{-1}\left(\frac{\sqrt{2 v}+1}{\sqrt{2}}\right)\right]

=\sqrt{2}\left[\tan ^{-1}\left(\frac{\sqrt{2 v}-1}{\sqrt{2}}\right)+\tan ^{-1}\left(\frac{\sqrt{2 v}+1}{\sqrt{2}}\right)\right] \quad[\because v=\sqrt{u}, u=\tan x]

=\sqrt{2}\left[\tan ^{-1}\left(\frac{(\sqrt{2 \tan x}-1)}{\sqrt{2}}\right)+\tan ^{-1}\left(\frac{\sqrt{2 \tan x}+1}{\sqrt{2}}\right)\right]

 

 

Posted by

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