Question

integrate this question
​

Answer 1

In this problem ,

we need to change both sinx and tanx

with half angle formula ,

sinx=(2tanx/2)/(1+tan^2(x/2))

or

tanx= (2tanx/2)/(1-tan^2(x/2))

Now, put the value in given question

then , we use substitution

so, Let tanx/2 =t

=sec^2x/2 dx=2dt

=(1+tan^2(x/2))dx=2dt

【sec^2x-tan^2x=1】

=dx= 2dt/1+t^2

now put the value ..

formula:

$\int \frac{1}{x} dx = log(x)$

$\int {x}^{y} dx = \frac{x ^{y + 1} }{y + 1}$

soln refers to the attachment

Answer 2

Let

dx

I =∫-------------------------

sinx + tanx

dx

=∫----------------------------

sinx

( sinx + ------------)

cosx

dx

=∫ --------------------------------------

( sinx cosx +sinx)/cosx

cosx dx

=∫----------------------------

sinx (1+ cosx)

sinx cosx

=∫-----------------------------dx

sin²x (1 +cosx)

sinx cosx

=∫--------------------------------dx

(1-cos²x)(1+cosx)

sinx cosx

=∫--------------------------------------------dx

(1+cosx)(1-cosx)(1+cosx)

sinx cosx

=∫--------------------------------------dx

(1+cosx)² (1-cosx)

let cosx=t

- sinx dx=dt

sinx dx =-dt

-dt

I=∫--------------------

(1+t)² (1-t)

dt

=-∫------------------------

(1+t)² (1-t)

Now we do partial fraction kindly see attachement for working of partial fraction

1 1/4 1/2 1/4

-----------------=-------- - --------- + ----------

(1+t)²(1-t) 1+t (1+t)² (1-t)

now

1 1 1 1 1

I=-1/4∫------- dt + ----∫----------dt- ----∫-------dt

1+t 2 (1+t)² 4 1-t

1 1 - 1 1

=- -----log(1+t) + ----( -------) - ------- log(1-t)

4 2 1+t (- 4)

+c

- 1 1

= ----------- + -----{ log(1-t) -log(1+t)}+c

2(1+t) 4

- 1 1 1-cosx

=----------------- +------ log(----------------- ) +c

2(1+cosx) 4 1+cosx

Hope it helps u

Thanks