Question

Verify each of the following answers for an indefinite integral by one or more of the methods suggestedabove.

1. $\int 2\sin \theta \cos \theta d\theta ={\sin }^2\theta$or $-{\cos }^2\theta$or $-\frac{1}{2}\cos 2\theta$ Hint: Use trig identities.

Short answer

It is proved that$\int 2\sin \theta \cos \theta d\theta ={\sin }^2\theta$.

Step-by-step solution

The substitution method

The$u-$substitution reserves the chain rule and it is used to solve integrals.

Itis a methodto obtain antiderivatives. Generally, formula for $u-$substitution is:

$\int f\left(g\right(x\left)\right)g\text{'}\left(x\right)dx=\int f\left(u\right)du$

Here, $u=g\left(x\right);du=g\text{'}\left(x\right)dx$.

Evaluate the trigonometric identity

The given trigonometric integral is$\int 2\sin \theta \cos \theta d\theta$.

Take the constant out by using the property $\int a.f\left(x\right)dx=a.\int f\left(x\right)dx$.

So, after using the above formula in the trigonometric identity, we get:

$\int 2\sin \theta \cos \theta d\theta =2.\int \sin \left(\theta \right)\cos \left(\theta \right)d\left(\theta \right)$

Applying substitution

Let $u=\sin \theta$, then $du=\cos \theta d\theta$.

After applying u- substitution method, we get:

role="math" localid="1653045028107" $$\begin{gathered} 2.\int \sin \theta \cos \theta d\theta =2.\int udu \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}=2.\frac{{\mathit{u}}^{\mathit{2}}}{\mathit{2}}[\mathrm{Applying}\mathrm{Power}\mathrm{Rule}] \end{gathered}$$

Now, substituting back $u=\sin \theta$, we get:

$u={\sin }^2\theta$

Simplification of the identity

On further simplification:

$2.\frac{{\sin }^2\theta }{2}={\sin }^2\theta$

Hence,$\int 2\sin \theta \cos \theta d\theta ={\sin }^2\theta$.