Question

$A:\int \frac{1}{5+4\sin x}dx=\frac{2}{3}{\tan }^{-1}\left(\frac{4+5\tan \left(x/2\right)}{3}\right)+c$
$R$: lf $a>0,a>b$, then $\int \frac{dx}{a+b\sin x}=$ $\frac{2}{\sqrt{a^2-b^2}}{\tan }^{-1}\left[\frac{b+a\tan \left(x/2\right)}{\sqrt{a^2-b^2}}\right]+c$

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true but R is not correct explanation of A
• C. A is true R is false
• D. A is false but R is true

Answer 1

Answer: (A)

Both A and R are true and R is the correct explanation of A

$\int \frac{dx}{a+b\sin x}a>b$
$=\int \frac{dx}{a+b\frac{2\tan \left(x/2\right)}{1+{\tan }^{2}x/2}}$
$=\int \frac{{\sec }^{2}x/2dx}{a+a{\tan }^{2}x/2+2b\tan \left(x/2\right)}$
Put $\tan \left(x/2\right)=t$
$\frac{1}{2}{\sec }^2\left(x/2\right)dx=dt$
$=\int \frac{2dt}{a+a{t}^2+2bt}$
$=\frac{2}{a}\int \frac{dt}{1+t^2+\frac{2b}{a}t+\frac{b^2}{a^2}-\frac{b^2}{a^2}}$
$=\frac{2}{a}\int \frac{dt}{{(t+\frac{b}{a})}^2+{\left(\frac{\sqrt{a^2-b^2}}{a}\right)}^2}$
$=\frac{2}{\sqrt{a^2-b^2}}{\tan }^{-1}\left[\frac{t+\frac{b}{a}}{\frac{\sqrt{a^2-b^2}}{a}}\right]+c$
$=\frac{2}{\sqrt{a^2-b^2}}{\tan }^{-1}\left[\frac{b+a\tan \left(x/2\right)}{\sqrt{a^2-b^2}}\right]+c$
$\int \frac{dx}{a+b\sin x}=\frac{2}{\sqrt{a^2-b^2}}{\tan }^{-1}\left[\frac{b+a\tan \left(x/2\right)}{\sqrt{a^2-b^2}}\right]+ca>b$
Hence, reason is correct
$A:\int \frac{1}{5+4\sin x}dx=\frac{2}{3}{\tan }^{-1}\left(\frac{4+5\tan \left(x/2\right)}{3}\right)+c$
Put $a=5,b=4(a>b)$
$\int \frac{1}{5+4\sin x}dx=\frac{2}{3}{\tan }^{-1}\left(\frac{4+5\tan \left(x/2\right)}{3}\right)+c$
Hence, assertion is correct and reason is the correct explanation for assertion