Question

what is the integral of this please help.
This is from integration chapter
Class 12 jee mains level
∫(cotx-tanx )dx

Answer 1

Answer:   $log(sinx) + log(secx+tanx)+c$

Step-by-step explanation:

$\int {(cotx-tanx)} \, dx$

$\int {cotx} \, dx -\int {tanx} \, dx$

$\int {\frac{cosx}{sinx} } \, dx - \int {tanx\frac{tanx+secx}{tanx+secx} } \, dx$

$\int {\frac{cosx}{sinx} } \, dx - \int {\frac{tan^{2} x+secx*tanx}{tanx+secx} } \, dx$      

we know that derivative of     $\frac{d(tanx+secx)}{dx}$   =  $tan^{2} x+tanx*secx$

and that of   $\frac{d(sinx)}{dx} =cosx$

Let  $sinx=t$    -----------> (1.)

=>  $(cox)dx=dt$  ----------> (2.)

and   $tanx+secx=u$  -----------> (3.)

=>     $(tan^{2} x+secx*tanx)dx=dt$  ------------> (4.)

put (1.) , (2.), (3.) and (4.) in the integral :

$\int {\frac{1}{t} } \, dt - \int {\frac{1}{u} } } \, du$                   {  $\int {\frac{1}{x} } \, dx = log(x)+c$  }

$log(t)+log(u)+c$

$log(sinx) + log(secx+tanx)+c$

Answer 2

Answer:

∫(cotx−tanx)dx

\int {cotx} \, dx -\int {tanx} \, dx∫cotxdx−∫tanxdx

\int {\frac{cosx}{sinx} } \, dx - \int {tanx\frac{tanx+secx}{tanx+secx} } \, dx∫

sinx

cosx

dx−∫tanx

tanx+secx

tanx+secx

dx

\int {\frac{cosx}{sinx} } \, dx - \int {\frac{tan^{2} x+secx*tanx}{tanx+secx} } \, dx∫

sinx

cosx

dx−∫

tanx+secx

tan

2

x+secx∗tanx

dx

we know that derivative of \frac{d(tanx+secx)}{dx}

dx

d(tanx+secx)

= tan^{2} x+tanx*secxtan

2

x+tanx∗secx

and that of \frac{d(sinx)}{dx} =cosx

dx

d(sinx)

=cosx

Let sinx=tsinx=t -----------> (1.)

=> (cox)dx=dt(cox)dx=dt ----------> (2.)

and tanx+secx=utanx+secx=u -----------> (3.)

=> (tan^{2} x+secx*tanx)dx=dt(tan

2

x+secx∗tanx)dx=dt ------------> (4.)

put (1.) , (2.), (3.) and (4.) in the integral :

\int {\frac{1}{t} } \, dt - \int {\frac{1}{u} } } \, du { \int {\frac{1}{x} } \, dx = log(x)+c∫

x

1

dx=log(x)+c }

log(t)+log(u)+clog(t)+log(u)+c

log(sinx) + log(secx+tanx)+clog(sinx)+log(secx+tanx)+c