- Thread starter
- #1

#### 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

∫(sinx/1+sinx)dx

∫(sin^2x cos^2x)dx

- Thread starter paulmdrdo
- Start date

- Thread starter
- #1

- May 13, 2013

- 386

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

∫(sinx/1+sinx)dx

∫(sin^2x cos^2x)dx

∫(sinx/1+sinx)dx

∫(sin^2x cos^2x)dx

- Admin
- #2

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 \(\displaystyle \sin^2(\theta)=\frac{1-\cos(2\theta)}{2}\).

- Mar 1, 2012

- 249

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

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

Using the Pythagorean Identity which

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

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

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