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Integration of trig. functions

paulmdrdo

Active member
May 13, 2013
386
I have no idea how to solve these. please help.

∫(sinx/1+sinx)dx

∫(sin^2x cos^2x)dx
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: integration of trig func.

I will offer some suggestions:

1.) \(\displaystyle \int\frac{\sin(x)}{1+\sin(x)}\,dx\)

Multiply the integrand by \(\displaystyle 1=\frac{1-\sin(x)}{1-\sin(x)}\). Then apply the Pythagorean identity \(\displaystyle 1-\sin^2(\theta)=\cos^2(\theta)\) on the denominator.

2.) \(\displaystyle \int\sin^2(x)\cos^2(x)\,dx\)

Try the double-angle identity for sine \(\displaystyle \sin(2\theta)=2\sin(\theta)\cos(\theta)\). Then try the power reduction identity for sine \(\displaystyle \sin^2(\theta)=\frac{1-\cos(2\theta)}{2}\).
 

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
Re: integration of trig func.

Then try the power reduction identity for sine \(\displaystyle \sin^2(\theta)=\frac{1-\cos(2\theta)}{2}\).
You [OP] can also reach this result by using the double angle identity for cos if you're unfamiliar with the power reduction identity. See spoiler for derivation.

Using the addition identity for cos: \(\displaystyle \cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)\)

\(\displaystyle \cos (2A) = \cos(A+A) = \cos^2(A) - \sin^2(A)\)

Using the Pythagorean Identity which MarkFL put in post 2: \(\displaystyle \cos^2(A) = 1 - \sin^2(A)\)

\(\displaystyle \cos(2A) = 1-\sin^2(A) - \sin^2(A) = 1-2\sin^2(A)\)

Now to rearrange it to make \(\displaystyle \sin(A)\) the subject:

\(\displaystyle \sin(A) = \dfrac{1-\cos(2A)}{2}\)