/*! 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 11 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 11

Evaluate the following integrals : $$ \int x^{34}\left(1+x^{78}\right)^{1 / 2} d x $$

Short Answer

Expert verified
Question: Evaluate the integral 鈭玿^34(1+x^78)^(1/2) dx Answer: The evaluated integral is 鈭玿^34(1+x^78)^(1/2) dx = (1+x^78)/(117) (1 + x^78)^(1/2) + C, where C is the constant of integration.

Step by step solution

01

Choose a substitution function

Let's make a substitution: $$ u = (1+x^{78})^{1/2} $$
02

Differentiate the substitution function

Now, differentiate \(u\) with respect to \(x\). We get: $$ \frac{d u}{d x} = \frac{1}{2}(1 + x^{78})^{-\frac{1}{2}} \cdot 78 x^{77} $$
03

Rearrange to find dx

We want to rearrange this equation to solve for \(dx\). In order to do that, we first need to express \(x^{77}\) in terms of \(u\). If we square both sides of our substitution, we get: $$ u^2 = 1 + x^{78} $$ Subtract 1 from both sides: $$ u^2 - 1 = x^{78} $$ Take the 77th root of both sides: $$ x^{77} = \sqrt[77]{u^2 - 1} $$ Now we can substitute this into our equation for \(du/dx\) and solve for \(dx\): $$ \frac{d u}{d x} = \frac{1}{2}(1 + x^{78})^{-\frac{1}{2}} \cdot 78 \sqrt[77]{u^2 - 1} $$ $$ \frac{d u}{d x} = \frac{1}{2} u^{-1} \cdot 78 \sqrt[77]{u^2 - 1} $$ Rearranging to solve for \(dx\): $$ dx = \frac{2}{78} \cdot u \cdot \frac{du}{\sqrt[77]{u^2 - 1}} $$
04

Substitute back into original integral

Now that we have our substitution and the expression for \(dx\), we can rewrite the original integral in terms of \(u\): $$ \int x^{34}\left(1+x^{78}\right)^{1 / 2} dx = \int \left(\sqrt[77]{u^2 - 1}\right)^{34} u \cdot \frac{2}{78} \cdot u \cdot \frac{d u}{\sqrt[77]{u^2 - 1}} $$
05

Simplify the integral

Now we can simplify the integral: $$ \frac{2}{78} \int u^2 du = \frac{u^3}{117} $$
06

Substitute back \(x\) for \(u\)

Now we will substitute back \(u = (1 + x^{78})^{1/2}\) to find our final answer: $$ \frac{u^3}{117} = \frac{\left((1 + x^{78})^{1/2}\right)^3}{117} = \frac{1+x^{78}}{117} (1 + x^{78})^{1 / 2} $$ So, the evaluated integral is: $$ \int x^{34}\left(1+x^{78}\right)^{1 / 2} dx = \frac{1+x^{78}}{117} (1 + x^{78})^{1 / 2} + C $$ Where, \(C\) is the constant of integration.

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Most popular questions from this chapter

Evaluate the following integrals: (i) \(\int \frac{2 x^{3}+3 x^{2}+4 x+5}{2 x+1} d x\) (ii) \(\int\left(\frac{x^{-6}-64}{4+2 x^{-1}+x^{-2}}, \frac{x^{2}}{4-4 x^{-1}+x^{-2}} \frac{4 x^{2}(2 x+1)}{1-2 x}\right) \mathrm{dx}\) (iii) \(\int\left(\frac{\sqrt{x}}{2}-\frac{1}{2 \sqrt{x}}\right)^{2}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right) d x\) (iv) \(\int \frac{\sqrt{1-x^{2}}+1}{\sqrt{1-x}+1 / \sqrt{1+x}} d x\).

Evaluate the following integrals: (i) \(\int \frac{2 x+\sin 2 x}{1+\cos 2 x} d x\) (ii) \(\int\left(\tan (\ln x)+\sec ^{2}(\ln x)\right\\} d x\) (iii) \(\int \frac{x+\sqrt{\left(1-x^{2}\right)} \sin ^{-1} x}{\sqrt{\left(1-x^{2}\right)}} d x\)

Evaluate the following integrals : $$ x\left(1+8 x^{3}\right)^{1 / 3} d x $$

Derive the reduction formula \(\int \cos ^{n} x d x=\frac{1}{n} \cos ^{n-1} x \sin x+\frac{n-1}{n} \int \cos ^{n-2} x d x\).

Evaluate the following integrals: $$ \int \frac{x^{2} d x}{\sqrt{1-2 x-x^{2}}} $$
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