Question

What is the integral of $\sin^5 x \cos^3 x \, dx$ if the lower limit is zero and the upper limit is $\pi/2$?

• A. 0.0203
• B. 0.0307
• C. 0.0417
• D. 0.0543

Answer 1

Answer: (C)

0.0417

Answer 2

To evaluate the definite integral $\int_0^{\pi/2} \sin^5 x \cos^3 x \, dx$, we can use substitution.

Let $u = \sin x$, then $du = \cos x \, dx$. Also, $\cos^2 x = 1 - \sin^2 x = 1 - u^2$.

The integral becomes: $\int_0^{\pi/2} \sin^5 x \cos^3 x \, dx = \int_0^{\pi/2} \sin^5 x \cos^2 x \cos x \, dx = \int_0^{\pi/2} \sin^5 x (1 - \sin^2 x) \cos x \, dx$

When $x = 0$, $u = \sin(0) = 0$. When $x = \pi/2$, $u = \sin(\pi/2) = 1$.

So, the integral in terms of $u$ is: $\int_0^1 u^5 (1 - u^2) \, du = \int_0^1 (u^5 - u^7) \, du$

Now, integrate term by term: $\int_0^1 (u^5 - u^7) \, du = \left[ \frac{u^6}{6} - \frac{u^8}{8} \right]_0^1$

Evaluate at the limits: $\left( \frac{1^6}{6} - \frac{1^8}{8} \right) - \left( \frac{0^6}{6} - \frac{0^8}{8} \right) = \frac{1}{6} - \frac{1}{8} = \frac{4}{24} - \frac{3}{24} = \frac{1}{24}$

Converting to decimal: $\frac{1}{24} \approx 0.041666...$

Rounding to four decimal places gives $0.0417$.