Question

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

• 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

$\int \frac{dx}{a+b\sin x}$
$=\int \frac{dx}{a+b\frac{2\tan \left(x/2\right)}{1+{\tan }^{2}x/2}}$
$=\int \frac{{\sec }^{2}x/2dx}{a+a{\tan }^{2}x/2+2b\tan \left(x/2\right)}$
Substituting $\tan \left(x/2\right)=t$
$\frac{1}{2}{\sec }^2\left(x/2\right)dx=dt$
$=\int \frac{2dt}{a+a{t}^2+2bt}$
$=\frac{2}{a}\int \frac{dt}{1+t^2+\frac{2b}{a}t+\frac{b^2}{a^2}-\frac{b^2}{a^2}}$
$=\frac{2}{a}\int \frac{dt}{{(t+\frac{b}{a})}^2-{\left(\frac{\sqrt{b^2-a^2}}{a}\right)}^2}$
$=\frac{1}{\sqrt{b^2-a^2}}\log |\frac{t+\frac{b}{a}-\frac{\sqrt{b^2-a^2}}{a}}{t+\frac{b}{a}+\frac{\sqrt{b^2-a^2}}{a}}|$
$=\frac{1}{\sqrt{b^2-a^2}}\log \left|\frac{a\tan \left(x/2\right)+b-\sqrt{b^2-a^2}}{a\tan \left(x/2\right)+b+\sqrt{b^2-a^2}}\right|+c$
$\int \frac{dx}{a+b\sin x}=\frac{1}{\sqrt{b^2-a^2}}\log \left|\frac{a\tan \left(x/2\right)+b-\sqrt{b^2-a^2}}{a\tan \left(x/2\right)+b+\sqrt{b^2-a^2}}\right|+c$
Thus, reason is correct .
$A:\int \frac{1}{4+5\sin x}dx=\frac{1}{3}\log \left|\frac{2\tan \left(x/2\right)+1}{2\tan \left(x/2\right)+4}\right|+c$
Substitute$a=4$,$b=5$
$\int \frac{1}{4+5\sin x}dx$ $=$ $\frac{1}{3}\log \left|\frac{2\tan \left(x/2\right)+1}{2\tan \left(x/2\right)+4}\right|+c$
Thus, assertion is true and reason is correct explanation for assertion
Hence, option 'A' is correct.