Question

Integrate tan^2(2x+5) dx

Answer 1

$\huge \mathfrak{solution}$

To find:

$\int \: {tan}^{2} (2x + 5) \: dx$

As we know that :

${ \sec }^{ ^{2} } \theta \: - { \tan}^{2} \theta = 1 \\ = > \tan ^{2} \theta = { \sec}^{2} \theta \: - 1$

$\star \int \: {sec}^{2} (2x + 5)dx - \int \: 1 \: dx$

Here :

Let

=> 2x+5 = t

=> 2dx= dt

=> dx= dt/2

$\star \int \: {sec}^{2} (t) \frac{dt}{2} - \int \: 1dx \\ \\ \star \: \frac{1}{2} \int {sec}^{2} t \: dt - \int \: 1dx \\ \\ \star \: \frac{1}{2} tan \: (t) - x + c$

Now , put the value of (t)

$\star \: \frac{1}{2} tan(2x + 5) - x + c$

Formula used :

$\leadsto \red{ \int}\boxed{ \bold { {x}^{n}dx = \frac{ {x}^{n + 1} }{n + 1}} + c }$

$\leadsto \boxed{ \bold{ \green{\int \: {sec}^{2} x \: dx \: = tan \: x \: + c}}}$

Answer 2

Answer:

plz find the attached file