/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 1

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

Expert verified
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

01

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.
02

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: $$ \int_0^{\frac{\pi}{2}}\sin^5{x} \space dx = \frac{1}{16}\left[10\left(-\frac{\cos{x}}{1}\right) - 5\left(-\frac{\cos{3x}}{3}\right) + \left(-\frac{\cos{5x}}{5}\right)\right]_0^{\frac{\pi}{2}} $$
03

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: $$ \int_0^{\frac{\pi}{2}}\sin^5{x} \space dx = \frac{16}{45} $$
04

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.
05

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: $$ \int_0^{\frac{1}{2}\pi}\cos^6{x} \space dx = \frac{1}{32}\left[\left(15x\right) + \left(\frac{45\sin{2x}}{2}\right) + \left(\frac{15\sin{4x}}{4}\right) + \left(\frac{\sin{6x}}{6}\right)\right]_0^{\frac{1}{2}\pi} $$
06

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: $$ \int_0^{\frac{1}{2}\pi}\cos^6{x} \space dx = \frac{15\pi}{64} $$

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Most popular questions from this chapter

Evaluate the integrals (i) \(\int_{0}^{b} \frac{x d x}{(1+x)^{3}}\) (ii) \(\int_{0}^{b} \frac{x^{2} d x}{(1+x)^{4}}\) and show that they converge to finite limits as \(\mathrm{b} \rightarrow \infty\)

Suppose that the velocity function of a particle moving along a line is \(v(t)=3 t^{3}+2\). Find the average velocity of the particle over the time interval \(1 \leq \mathrm{t} \leq 4\) by integrating.

Which of following integrals are improper ? Why? (a) \(\int_{1}^{2} \frac{1}{2 x-1} \mathrm{dx}\) (b) \(\int_{0}^{1} \frac{1}{2 x-1} d x\) (c) \(\int_{-\infty}^{\infty} \frac{\sin x}{1+x^{2}} d x\) (d) \(\int_{1}^{2} \ln (x-1) d x\)

Let \(\mathrm{P}_{\mathrm{n}}\) denote the polynomial of degree \(\mathrm{n}\) given by \(\mathrm{P}_{\mathrm{n}}(\mathrm{x})=\mathrm{x}+\frac{\mathrm{x}^{2}}{2}+\frac{\mathrm{x}^{3}}{3}+\ldots .+\frac{\mathrm{x}^{\mathrm{n}}}{\mathrm{n}}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{\mathrm{x}^{\mathrm{k}}}{\mathrm{k}}\). Then, for every \(x<1\) and every \(n \geq 1\), prove that \(-\ln (1-x)=P_{n}(x)+\int_{0}^{x} \frac{u^{n}}{1-u} d u\)

Compute (a) \(\lim _{t \rightarrow 0+} \int_{t}^{1} \frac{1}{x} \mathrm{dx}\) (b) \(\lim _{t \rightarrow 1-} \int_{0}^{t} \tan \frac{\pi}{2} x d x\). How does the result give insight into the fact that neither integrand is integrable over the interval \([0,1] ?\)
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