Q.Evaluate the integral using substitution $\int_{0}^{\pi/2}\sqrt{\sin\phi}\,\cos^5\phi\,d\phi$

An odd power of $\cos$ sits next to $\sqrt{\sin\phi}$, so substitute $t=\sin\phi$; one $\cos\phi$ becomes $dt$ and the rest turns into a polynomial.

Let $t=\sin\phi$, $dt=\cos\phi\,d\phi$. Then $\cos^5\phi\,d\phi=(1-\sin^2\phi)^2\cos\phi\,d\phi=(1-t^2)^2\,dt$, and $\phi:0\to\tfrac\pi2$ gives $t:0\to1$:

With common denominator $231$: $\dfrac{154-132+42}{231}=\dfrac{64}{231}$.

Choosing the substitution

Because $\cos\phi$ appears to an odd power, we can peel off one factor to serve as $dt$ and convert the even remainder to $\sin\phi$. Set $t=\sin\phi$, so $dt=\cos\phi\,d\phi$.

1. Rewrite the integrand

The limits change as $\phi:0\to\tfrac\pi2$ gives $t:0\to1$, so

2. Expand and integrate

3. Add the fractions

Common denominator $231$:

Method: Odd power of sine/cosine — peel off one factor for the differential

When one trig function appears to an odd power, split off a single factor to become $du$ and convert the remaining even power using $\sin^2+\cos^2=1$.

Steps

Step 1: Locate the odd power and pick the other function as $u$.

If $\cos x$ has an odd power, set $u=\sin x$ (so $du=\cos x\,dx$); if $\sin x$ is odd, set $u=\cos x$.

Step 2: Reserve one factor for $du$ and rewrite the rest.

Write $\cos^{2k+1}x = (\cos^2x)^k\cos x = (1-\sin^2x)^k\cos x$, turning the even remainder into a polynomial in $u$.

Step 3: Change the limits and integrate the polynomial.

Convert $x$-limits to $u$-limits and integrate term by term with the power rule $\displaystyle\int u^n\,du=\frac{u^{n+1}}{n+1}$.

Step 4: Sum the fractional-power terms over a common denominator to finish.

Common Mistakes

Mistake 1: Substituting $t=\sin\phi$ but forgetting to convert the even remaining cosines.

Why it's wrong: after peeling one $\cos\phi$ for $dt$, the leftover $\cos^4\phi$ must become $(1-\sin^2\phi)^2=(1-t^2)^2$; leaving a stray $\cos$ or $\phi$ makes the integral unintegrable in $t$. Correct approach: use $\cos^2\phi=1-\sin^2\phi$ on the even part.

Mistake 2: Mishandling fractional exponents when integrating.

Why it's wrong: $\int t^{1/2}\,dt=\tfrac23 t^{3/2}$ and $\int t^{9/2}\,dt=\tfrac{2}{11}t^{11/2}$ — adding $1$ to a half-integer power trips students up. Correct approach: apply the power rule carefully to each term and add over the common denominator $231$.