Question

Reduction formulas can be used to compute integrals of higher power of sin x, cos x, tan x etc.

$\int sin^5 x dx = -\frac{1}{5}sin^4 x cos x + A sin^2 x cos x - \frac{8}{15}cos x + C$, then A equals:

• A. $-\frac{2}{15}$
• B. $-\frac{3}{5}$
• C. $-\frac{4}{15}$
• D. $-\frac{1}{15}$

Answer 1

Answer: (C)

$-\frac{4}{15}$

Answer 2

The reduction formula for the integral of $\sin^n x$ is given by: $\int \sin^n x \, dx = -\frac{1}{n} \sin^{n-1} x \cos x + \frac{n-1}{n} \int \sin^{n-2} x \, dx$ For $n=5$: $\int \sin^5 x \, dx = -\frac{1}{5} \sin^{5-1} x \cos x + \frac{5-1}{5} \int \sin^{5-2} x \, dx$ $\int \sin^5 x \, dx = -\frac{1}{5} \sin^4 x \cos x + \frac{4}{5} \int \sin^3 x \, dx$ Now, we apply the reduction formula for $n=3$: $\int \sin^3 x \, dx = -\frac{1}{3} \sin^{3-1} x \cos x + \frac{3-1}{3} \int \sin^{3-2} x \, dx$ $\int \sin^3 x \, dx = -\frac{1}{3} \sin^2 x \cos x + \frac{2}{3} \int \sin x \, dx$ $\int \sin^3 x \, dx = -\frac{1}{3} \sin^2 x \cos x + \frac{2}{3} (-\cos x) + C'$ $\int \sin^3 x \, dx = -\frac{1}{3} \sin^2 x \cos x - \frac{2}{3} \cos x + C'$ Substitute this result back into the expression for $\int \sin^5 x \, dx$: $\int \sin^5 x \, dx = -\frac{1}{5} \sin^4 x \cos x + \frac{4}{5} \left( -\frac{1}{3} \sin^2 x \cos x - \frac{2}{3} \cos x \right) + C$ $\int \sin^5 x \, dx = -\frac{1}{5} \sin^4 x \cos x - \frac{4}{15} \sin^2 x \cos x - \frac{8}{15} \cos x + C$ The given expression is: $\int \sin^5 x \, dx = -\frac{1}{5}\sin^4 x \cos x + A \sin^2 x \cos x - \frac{8}{15}\cos x + C$ Comparing the derived expression with the given expression, the coefficient of the $\sin^2 x \cos x$ term is $A$. Therefore, $A = -\frac{4}{15}$.