Integration-EASY!!!!!!!!!!!

8 Answers

1
Siddharth ·

The question is possible if it is sin4x-4
Or else the question is not possible
Answer:
∫ex(sin4x dx)/1-cos4x -∫2exdx/1-cos4x

U can write 1-cos4x as 2sin22x and sin4x as 2sin2x.cos4x
∫excot2xdx -∫exdx/sin22x
∫(excot2x dx-excosec22x)dx
now this in the form of ∫ex(f(x)+f1(x)=ex.f(x)

Now the answer is excot2x....

1
ith_power ·

why isn't the question possible if 2 is given in place of 4 :) :)

62
Lokesh Verma ·

dude the quesiton is possible.. but i am not sure if we have an answer at hand..

if u do it will be gr8 :)

1
ith_power ·

if 4 is given in place of 2 then we can write it as e^x(f(x)+f'(x)) where f(x)= sin4x/(1-cos 4x) :)

62
Lokesh Verma ·

This can be done if and only if we can solve

∫excot2x dx

anyone game?

1
Siddharth ·

I have tried... But i believe something is wrong.. I have a strong feeling...
I think ∫excot2x dx can be done using ILATE rule... I have not tried it yet...

1
ith_power ·

i think original question was e^2x

given term= ∫e^2x(cot 2x - (cosec 2x)^2) dx
integrating by parts,

cot 2x * (e^ 2x)/2 - ∫(-2)cosec22x e2x/2 dx - ∫e^2x cosec22x.
= cot 2x * (e^ 2x)/2 +c

1
ith_power ·

I KNOW SOLUTION IS WEIRD BUT ITS 100% TRUE..............
WHERE 2F1 IS HYPERGEOMETRIC FUNCTION..
REF. http://reference.wolfram.com/mathematica/ref/Hypergeometric2F1.html

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