Find integration of logs sin dx whose upper and lower limit are pi and zero
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.
Answer laterif i=integral (e^x) log(e^x+1)
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!!!
find the integration of square root of tax
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
