How Do You Integrate These Trigonometric Functions?

[paulmdrdo1]

I have no idea how to solve these. please help.

∫(sinx/1+sinx)dx

∫(sin^2x cos^2x)dx

 

[MarkFL]

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]

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.

Spoiler

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:

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