Integrate the function: $\frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}}$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Let $\tan x = t$.
Then,$\sec ^{2} x \, dx = dt$.
Substituting these into the integral,we get:
$\int \frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}} \, dx = \int \frac{dt}{\sqrt{t^{2} + 2^{2}}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{x^{2} + a^{2}}} = \log |x + \sqrt{x^{2} + a^{2}}| + C$,we have:
$= \log |t + \sqrt{t^{2} + 4}| + C$.
Substituting back $t = \tan x$,the final result is:
$= \log |\tan x + \sqrt{\tan^{2} x + 4}| + C$,where $C$ is an arbitrary constant.

Explore More

Similar Questions

If $\int \frac{2 \sin 2x - 3 \cos x}{2 \sin^2 x - 3 \sin x + 4} dx = f(x) + c$ where $c$ is the constant of integration,then $f\left(\frac{\pi}{2}\right) - f(0) =$

$\int \frac{e^{2x}}{\sqrt[4]{e^x+1}} dx =$

For $x < 1$,evaluate $\int \frac{x-x^2}{\sqrt{1-x}} d x$.

To evaluate $\int x^3 e^{3x^2 + 5} dx$,the simplest way is to

$f(x) = \int \frac{dx}{\sin^6 x}$ is a polynomial of degree

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo