Question

$A:\int \frac{1}{3+4\cos x}dx=\frac{1}{\sqrt{7}}log\frac{\sqrt{7}+\tan \left(x/2\right)}{\sqrt{7}-\tan \left(x/2\right)}+c$
$R:$ If $a<b$ then $\int \frac{dx}{a+b\cos x}=\frac{1}{\sqrt{b^2-a^2}}\log \left|\frac{\sqrt{b+a}+\sqrt{b-a}\tan \left(x/2\right)}{\sqrt{b+a}-\sqrt{b-a}\tan \left(x/2\right)}\right|+c$

• 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

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

$\int \frac{dx}{a+b\cos x}=\frac{1}{\sqrt{b^2-a^2}}\log \left|\frac{\sqrt{b+a}+\sqrt{b-a}\tan \left(x/2\right)}{\sqrt{b+a}-\sqrt{b-a}\tan \left(x/2\right)}\right|+c$
$\int \frac{dx}{a+b\cos x}$
$=\int \frac{dx}{a+b\left(\frac{1-{\tan }^{2}x/2}{1+{\tan }^{2}x/2}\right)}$
$=\int \frac{{\sec }^{2}x/2dx}{(a+b)+(a-b){\tan }^{2}x/2}$
Substituting $\tan \left(x/2\right)=t$
$\frac{1}{2}{\sec }^2\left(x/2\right)dx=dt$
$=\int \frac{2dt}{(a+b)-(b-a)t^2}$
$=\int \frac{2dt}{{(b-a)\left[\right(\sqrt{\frac{b+a}{b-a}})}^2-t^2]}$
$=\frac{1}{\sqrt{b^2-a^2}}\log \left|\frac{\sqrt{b+a}+\sqrt{b-a}\tan \left(x/2\right)}{\sqrt{b+a}-\sqrt{b-a}\tan \left(x/2\right)}\right|+c$ ....................for $(a<b)$
Hence, reason is true.
$A:\int \frac{1}{3+4\cos x}dx=\frac{1}{\sqrt{7}}log\frac{\sqrt{7}+\tan \left(x/2\right)}{\sqrt{7}-\tan \left(x/2\right)}+c$
For assertion, Substituting $a=3$, $b=4$, (a$\int \frac{1}{3+4\cos x}dx=\frac{1}{\sqrt{7}}log\frac{\sqrt{7}+\tan \left(x/2\right)}{\sqrt{7}-\tan \left(x/2\right)}+c$
Therefore, assertion is true and reason is correct explanation for assertion
Hence, option 'A' is correct.