Question

Evaluate the following integrals : (i) \(\int_{0}^{\pi / 2} \sin ^{5} x d x\) (ii) \(\int_{0}^{\frac{1}{2} \pi} \cos ^{6} x d x\)

Short answer

Compare the results of the following definite integrals: 1) \(\int_0^{\frac{\pi}{2}}\sin^5{x} \space dx\) 2) \(\int_0^{\frac{\pi}{2}}\cos^6{x} \space dx\) Answer: 1) The definite integral \(\int_0^{\frac{\pi}{2}}\sin^5{x} \space dx\) evaluates to \(\frac{16}{45}\). 2) The definite integral \(\int_0^{\frac{\pi}{2}}\cos^6{x} \space dx\) evaluates to \(\frac{15\pi}{64}\). Though both integrals involve odd and even powers of trigonometric functions and are evaluated on the same interval, their values are different. The first integral has a rational value, while the second has an irrational value involving \(\pi\).

Step-by-step solution

Applying the power-reduction formula for sin^5{x}

The power-reduction formula for sin^5{x} is given by: $$ \sin^5{x} = \frac{1}{16}\left(10\sin{x} - 5\sin{3x} + \sin{5x}\right) $$ Now we will integrate this expression with respect to x in the given interval.

Integrating the power-reduced function for sin^5{x}

To integrate the power-reduced function for sin^5{x}, we integrate each term in the sum separately: $$ \int_0^{\frac{\pi}{2}}\sin^5{x} \space dx = \frac{1}{16}\left(\int_0^{\frac{\pi}{2}}10\sin{x} \space dx - \int_0^{\frac{\pi}{2}}5\sin{3x} \space dx + \int_0^{\frac{\pi}{2}}\sin{5x} \space dx \right) $$ Using the antiderivative rules for sums and the antiderivative of sin(ax), we get: $$ \int_0^{\frac{\pi}{2}}\sin^5{x} \space dx = \frac{1}{16}\left[10\left(-\frac{\cos{x}}{1}\right) - 5\left(-\frac{\cos{3x}}{3}\right) + \left(-\frac{\cos{5x}}{5}\right)\right]_0^{\frac{\pi}{2}} $$

Applying the limits for sin^5{x}

Now, apply the limits of integration and use the trigonometric identities for the sine and cosine functions to get the final answer for the first integral: $$ \int_0^{\frac{\pi}{2}}\sin^5{x} \space dx = \frac{1}{16}\left[10\left(-\frac{\cos{\frac{\pi}{2}}}{1} + \frac{\cos{0}}{1}\right) - 5\left(-\frac{\cos{\frac{3\pi}{2}}}{3} + \frac{\cos{0}}{3}\right) + \left(-\frac{\cos{\frac{5\pi}{2}}}{5} + \frac{\cos{0}}{5}\right)\right] = \frac{1}{16}(10 - \frac{5}{3} + \frac{1}{5}) $$ Thus, we have: $$ \int_0^{\frac{\pi}{2}}\sin^5{x} \space dx = \frac{16}{45} $$

Applying the power-reduction formula for cos^6{x}

The power-reduction formula for cos^6{x} is given by: $$ \cos^6{x} = \frac{1}{32}\left(15 + 45\cos{2x} + 15\cos{4x} + \cos{6x}\right) $$ Now we will integrate this expression with respect to x in the given interval.

Integrating the power-reduced function for cos^6{x}

To integrate the power-reduced function for cos^6{x}, we integrate each term in the sum separately: $$ \int_0^{\frac{1}{2}\pi}\cos^6{x} \space dx = \frac{1}{32}\left(\int_0^{\frac{1}{2}\pi}15 \space dx + \int_0^{\frac{1}{2}\pi}45\cos{2x} \space dx + \int_0^{\frac{1}{2}\pi}15\cos{4x} \space dx + \int_0^{\frac{1}{2}\pi}\cos{6x} \space dx\right) $$ Using the antiderivative rules for sums and the antiderivative of cos(ax), we get: $$ \int_0^{\frac{1}{2}\pi}\cos^6{x} \space dx = \frac{1}{32}\left[\left(15x\right) + \left(\frac{45\sin{2x}}{2}\right) + \left(\frac{15\sin{4x}}{4}\right) + \left(\frac{\sin{6x}}{6}\right)\right]_0^{\frac{1}{2}\pi} $$

Applying the limits for cos^6{x}

Now, apply the limits of integration and use the trigonometric identities for the sine and cosine functions to get the final answer for the second integral: $$ \int_0^{\frac{1}{2}\pi}\cos^6{x} \space dx = \frac{1}{32}\left[\left(15\cdot\frac{1}{2}\pi\right) + \left(\frac{45\sin{\pi}}{2}\right) - \left(\frac{15\sin{2\pi}}{4}\right) - \left(\frac{\sin{3\pi}}{6}\right) \right] = \frac{1}{32}(15\frac{\pi}{2} - 0 - 0 - 0) $$ Thus, we have: $$ \int_0^{\frac{1}{2}\pi}\cos^6{x} \space dx = \frac{15\pi}{64} $$