Hello,

We have to find the integration of logs. sinx.dx,

Intg.( logs.sinx.dx) = clearly logs is a constant, so take it out of the integration.

= logs. intg.( sinxdx) = - logs. cosx , now we have to put the upper and lower limits as pi and 0.

So we get,

- logs. ( -1 - 1 )

= 2 .logs

Hence this is the answer of the integration.

Hope it helps you.

Okh to solve your question I will use

i for integration and d for diffrentiation w.r.t. x okh

Apply ILATE rule

which says i= log(e^x+1) i e^xdx- i { d (log(e^x+1)) i e^xdx}dx

log(e^x+1)e^x- i {e^x/(e^x+1)*e^x}dx ( i e^x=e^x && d logx=1/x)

Now use substitution

[Let u=e^x then du=e^xdx]

The equation becomes

log(e^x+1)e^x- i {u/1+u}du

log(e^x+1)e^x- i {u+1-1/1+u}du

log(e^x+1)e^x- [ i du - i (1/1+u)du]

log(e^x+1)e^x- [u -log(1+u)] [ i ( 1/1+x)dx=log(1+x)]

Put u=e^x

log(e^x+1)e^x- [e^x -log(1+e^x)]

Hope this helps!!!

Hello aspirant,

√(tan x) dx

Let tan x = t 2

⇒ sec 2 x dx = 2t dt

⇒ dx = [2t / (1 + t 4 )]dt

⇒ Integral  ∫ 2t 2 / (1 + t 4 ) dt

⇒ ∫[(t 2 + 1) + (t 2 - 1)] / (1 + t 4 ) dt

⇒ ∫(t 2 + 1) / (1 + t 4 ) dt + ∫(t 2 - 1) / (1 + t 4 ) dt

⇒ ∫(1 + 1/t 2 ) / (t 2 + 1/t 2 ) dt + ∫(1 - 1/t 2 ) / (t 2 + 1/t 2 ) dt

⇒ ∫(1 + 1/t 2 )dt / [(t - 1/t) 2 + 2] + ∫(1 - 1/t 2 )dt / [(t + 1/t) 2 -2]

Let t - 1/t = u for the first integral ⇒ (1 + 1/t 2 )dt = du

and t + 1/t = v for the 2nd integral ⇒ (1 - 1/t 2 )dt = dv

Integral
= ∫du/(u 2 + 2) + ∫dv/(v 2 - 2)

= (1/√2) tan -1 (u/√2) + (1/2√2) log(v -√2)/(v + √2)l + c

= (1/√2) tan -1 [(t 2 - 1)/t√2] + (1/2√2) log (t 2 + 1 - t√2) / t 2 + 1 + t√2) + c

= (1/√2) tan -1 [(tanx - 1)/(√2tan x)] + (1/2√2) log [tanx + 1 - √(2tan x)] / [tan x + 1 + √(2tan x)] + c