what is the integration of sin^7 x. dx No Spam
Answers
Let me propose two different methods.
Method I: Reduction Formula
Let’s consider the general case of integrating sinnx w.r.t. x , where n>0 . Let
In=∫sinnxdx=∫sinn−1x⋅sinxdx
(integrate by parts)
=sinn−1x⋅(−cosx)−∫(n−1)sinn−2xcosx⋅(−cosx)dx
=−sinn−1xcosx+(n−1)⋅∫sinn−2xcos2xdx
=−sinn−1xcosx+(n−1)⋅∫sinn−2x(1−sin2x)dx
=−sinn−1xcosx+(n−1)⋅∫sinn−2xdx−(n−1)⋅∫sinnxdx
=−sinn−1xcosx+(n−1)⋅∫sinn−2xdx−(n−1)In
Then, In+(n−1)In=−sinn−1xcosx+(n−1)⋅∫sinn−2xdx
nIn=−sinn−1xcosx+(n−1)In−2
Therefore, In=−sinn−1xcosxn+(n−1)nIn−2
The aforementioned equation is called the reduction formula for ∫sinnxdx. Here's why. With n=7 , we obtain
∫sin7xdx=I7=−sin6xcosx7+67I5
=−sin6xcosx7+67⋅[−sin4xcosx5+45I3]
=−sin6xcosx7−6sin4xcosx35+2435I3
=−sin6xcosx7−6sin4xcosx35+2435[−sin2xcosx3+23I1]
=−sin6xcosx7−6sin4xcosx35−8sin2xcosx35+1635I1
But I1=∫sinxdx=−cosx+C . Therefore, taking c=16C35 , we obtain
I7=−cosx35(5sin6x+6sin4x+8sin2x+16)+c.
Method II: Complex Numbers
We have
sin7x=(eix−e−ix2i)7
(expand the numerator by the Binomial Theorem)
=e7ix−7e5ix+21e3ix−35eix+35e−ix−21e−3ix+7e−5ix−e−7ix−128i
=−164[e7ix−e−7ix2i−7⋅e5ix−e−5ix2i+21⋅e3ix−e−3ix2i−35⋅eix−e−ix2i]
=−164(sin7x−7sin5x+21sin3x−35sinx)
Therefore, ∫sin7xdx=164(cos7x7−7cos5x5+7cos3x−35cosx)+c
hope this ans. will help you dear
Answer:
sin^7 (x) dx = ∫ sin ^6 (x) sin(x) dx. Let u = cos(x), du = -sin( x) dx = ∫ {1 - cos²(x)}³ sin(x) dx. ➡️ (1/7)cos7(x) - (3/5)cos5(x) + cos3(x) - cos(x) + C.