Q.Integrate the following function: $\sin^3 x \cos^3 x$

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 $=2\int\cos^2 x\sin x\,dx=-2\int u^2\,du=-\tfrac23 u^3+C$. …

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

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 …

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

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

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

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 $0$. …

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

Therefore …

The integrand: …

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, …

Therefore …

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

$\int_0^{\pi/2}\sin2x\,e^{\sin x}\,dx=\int_0^1 2u\,e^u\,du.$ …

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

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