Q.Integrate the following function: $\sin 2x$

Apply the double-angle identity $\sin 2x=2\sin x\cos x$:

Therefore

Recall $\sin 2x=2\sin x\cos x$ and $\sin^2x+\cos^2x=1$, so

Taking the square root (on the principal branch where $\sin x+\cos x\ge 0$):

Integrate:

Concept and Intuition

Use the power-reduction identity $\sin^2\theta=\dfrac{1-\cos2\theta}{2}$.

Step-by-Step Solution

1. $\sin^2\pi x=\dfrac{1-\cos2\pi x}{2}$.
2. Integrate: $\dfrac12\int(1-\cos2\pi x)\,dx=\dfrac{x}{2}-\dfrac12\cdot\dfrac{\sin2\pi x}{2\pi}$.
3. This gives $\dfrac{x}{2}-\dfrac{1}{4\pi}\sin2\pi x+C$.

Common Mistakes

• Forgetting the $2\pi$ factor when integrating $\cos2\pi x$.
• Writing $\cos$ instead of $\sin$ in the result.

ANSWER: A

Therefore

The integrand:

Therefore

Concept and Intuition

Expanding $\sin 2x$ exposes a $\sin x$ factor to serve as $du$ when substituting $u=\cos x$.

Step-by-Step Solution

1. $\sin 2x\cos x=2\sin x\cos x\cdot\cos x=2\sin x\cos^2 x$.
2. Let $u=\cos x$, $du=-\sin x\,dx$.
3. Integral $=2\int\cos^2 x\sin x\,dx=-2\int u^2\,du=-\tfrac23 u^3+C$.
4. $=-\dfrac23\cos^3 x+C$.

Common Mistakes

• Forgetting the factor 2 from $\sin 2x=2\sin x\cos x$.

ANSWER: B

valid since $\cos x\ge 0$ on $[0,\pi/2]$. Let $u=\sin x$, $du=\cos x\,dx$:

Write $\sin2x=2\sin x\cos x$ and let $u=\sin x$, $du=\cos x\,dx$. Limits: $x=0\Rightarrow u=0$, $x=\tfrac{\pi}{2}\Rightarrow u=1$.

Integrate by parts ($\int u e^u du=(u-1)e^u$):

$1+\sin 2x=\sin^2 x+\cos^2 x+2\sin x\cos x=(\sin x+\cos x)^2$.

On $[0,\pi/4]$, $\sin x+\cos x\ge 0$, so $\sqrt{1+\sin 2x}=\sin x+\cos x$.

$\int_0^{\pi/4}(\sin x+\cos x)\,dx=[-\cos x+\sin x]_0^{\pi/4}=\big(\tfrac{\sqrt2}{2}-\tfrac{\sqrt2}{2}\big)-\big(0-1\big)=0+1=1$.

Concept and Intuition

Use the product-to-sum identity $\cos A\cos B=\tfrac12[\cos(A+B)+\cos(A-B)]$ before integrating.

Step-by-Step Solution

1. $\cos6x\cos2x=\tfrac12(\cos8x+\cos4x)$.
2. Antiderivative $=\tfrac12\left(\frac{\sin8x}{8}+\frac{\sin4x}{4}\right)$.
3. At $x=\pi/16$: $8x=\pi/2\Rightarrow\sin=1$; $4x=\pi/4\Rightarrow\sin=\frac{\sqrt2}{2}$. At $0$ all terms are $0$.
4. $=\tfrac12\left(\frac{1}{8}+\frac{\sqrt2/2}{4}\right)=\tfrac12\left(\frac18+\frac{\sqrt2}{8}\right)=\tfrac12\cdot\frac{1+\sqrt2}{8}=\frac{1+\sqrt2}{16}$.

Common Mistakes

• Forgetting the $\tfrac12$ from the product-to-sum step.
• Mis-evaluating $\sin(\pi/4)=\frac{\sqrt2}{2}$.

ANSWER: A

$\tan\dfrac{\theta}{2}\sin\theta\cos\theta=\dfrac{1-\cos\theta}{\sin\theta}\cdot\sin\theta\cos\theta=(1-\cos\theta)\cos\theta=\cos\theta-\cos^2\theta$.

$\displaystyle\int(\cos\theta-\cos^2\theta)\,d\theta=\sin\theta-\left(\dfrac{\theta}{2}+\dfrac{\sin2\theta}{4}\right)+C=\sin\theta-\dfrac{\theta}{2}-\dfrac{\sin2\theta}{4}+C$.

We use the identity $1+\sin\theta=\left(\sin\tfrac{\theta}{2}+\cos\tfrac{\theta}{2}\right)^2$.

Here $\theta=\tfrac{x}{8}$, so $\tfrac{\theta}{2}=\tfrac{x}{16}$ and

Taking the principal branch,

Thus the result is $16\sin\tfrac{x}{16}-16\cos\tfrac{x}{16}+C$.