Integrating Question with U-Sub: $\int\frac{\sin^3x}{\sqrt{\cos x}}\ dx$

Question

Question:

Find the value of the expression $$\int\frac{\sin^3x}{\sqrt{\cos x}}\ dx$$ by using the substitution $u=\cos x$.

My Working:

If $u=\cos x$, then

\begin{align} \frac{du}{dx}&=-\sin x\\ du&=-\sin x\cdot dx\\ \int\frac{\sin^3x}{\sqrt{\cos x}}\ dx&=-\int\frac{\sin^2x}{\sqrt{\cos x}}\cdot(-\sin x\cdot dx)\\ &=-\int\frac{1-\cos^2x}{\sqrt{\cos x}}\cdot(-\sin x\cdot dx)\\ &=-\int\frac{1-u^2}{u^{-\frac{1}{2}}}\cdot du\\ &=-\int u^{\frac{1}{2}}-u^{\frac{5}{2}}\cdot du\\ &=-\frac{2}{3}u^\frac{3}{2}+\frac{2}{7}u^{\frac{7}{2}}+c\\ &=-\frac{2}{3}(\cos x)^\frac{3}{2}+\frac{2}{7}(\cos x)^{\frac{7}{2}}+c \end{align}

However, my teacher says that I got this question wrong, and that the answer was

$$-\frac{2}{3}(\cos x)^\frac{3}{2}-\frac{2}{7}(\cos x)^{\frac{7}{2}}+c$$ and that I made a mistake. Could anyone please tell me why I got the question wrong? Thanks!

Answer 1

You've made mistake while substituting in terms of $u$.

$\begin{align}-\int\frac{\sin^2x}{\sqrt{\cos x}} (-\sin xdx) &=-\int\frac{1-u^2}{\color{red}{u^{+1/2}}}du = -\int(u^{-1/2} - u^{3/2})du\\ &=-\frac{1}{1/2} u^{1/2} + \frac{1}{5/2} u^{5/2} + c\\\\ & = -2 (\cos x)^{1/2} +\frac25 (\cos x)^{5/2}+c\end{align}$

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If the question were $\begin{align}\int{\sin^3x}\sqrt{\cos x} dx\end{align}$, then the result you've got is correct

$\begin{align}- \int{\sin^2x}\sqrt{\cos x} (-\sin xdx)&=-\int(1-u^2)u^{1/2} du = -\int (u^{1/2} - u^{5/2})du \\&= -\frac23 u^{3/2} + \frac27u^{7/2} + c\\\\ & = -\frac23 (\cos x)^{3/2} +\frac27 (\cos x)^{7/2} + c \end{align}$

Answer 2

Using integration by parts, we simplify the evaluation as $$ \begin{aligned} \int \frac{\sin ^3 x}{\sqrt{\cos x}} d x&=-2 \int \sin ^2 x d(\sqrt{\cos x}) \\ &=-2 \sin ^2 x \sqrt{\cos x}+4 \int \sin x \cos x \sqrt{\cos x} d x \\ &=-2 \sin ^2 x \sqrt{\cos x}-4 \int \cos ^{\frac{3}{2}} x d(\cos x) \\ &=-2 \sin ^2 x \sqrt{\cos x}-\frac{8}{5} \cos ^{\frac{5}{2}} x+C \end{aligned} $$