/*! 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 35 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 35

Evaluate the following integrals. $$\int \tan ^{3} 4 x d x$$

Short Answer

Expert verified
Question: Evaluate the integral \(\int \tan^{3}{4x} dx\). Answer: \(\int \tan^{3}{4x} dx = \tan^3{4x} \arctan{(\tan{4x})} + C\)

Step by step solution

01

Substitution

Let \(u = \tan{4x}\). Therefore, \(du = 4\sec^2{4x} dx\). Since \(\sec^2 4x = 1 + \tan^2 4x\), we have \(4 dx = du /(1 + u^2)\). Now, the original integration becomes: $$\int \tan ^{3} {4x} d x=\int u^3 \frac{du}{1+u^2}$$
02

Integration by parts

Use the integration by parts formula: \(\int u dv = uv - \int v du\). Let \(u_1 = u^3\) and \(dv_1 = \frac{du}{1+u^2}\). Then, \(du_1 = 3u^2 du\) and \(v_1 = \arctan{u}\). Applying the integration by parts formula, we get: $$\int u^3 \frac{du}{1+u^2} = u^3 \arctan{u} - 3 \int u^2 \arctan{u} du$$
03

Integration by parts (again)

To evaluate the integral \(\int u^2 \arctan{u} du\), we need to use integration by parts again. Set \(u_2 = \arctan{u}\) and \(dv_2 = u^2 du\). Therefore, \(du_2 = \frac{du}{1+u^2}\) and \(v_2 = \frac{1}{3}u^3\). By applying the integration by parts formula, we get: $$\int u^2 \arctan{u} du = \frac{1}{3} u^3 \arctan{u} - \frac{1}{3} \int u^2 \frac{du}{1+u^2}$$
04

Evaluating the integral

From the previous step, we have: $$\int u^3 \frac{du}{1+u^2} = u^3\arctan{u} - 3\left(\frac{1}{3}u^3 \arctan{u} - \frac{1}{3}\int u^2 \frac{du}{1+u^2}\right)$$ Simplifying the expression: $$\int u^3 \frac{du}{1+u^2} = u^3 \arctan{u} - u^3 \arctan{u} + \int u^2 \frac{du}{1+u^2}$$ Now, solving for the remaining integral, $$\int u^2 \frac{du}{1+u^2} = u^3 \arctan{u}$$
05

Reversing the substitution

Recall that we set \(u=\tan{4x}\). Substitute it back into the expression we obtained in the previous step: $$\int \tan^{3}{4x} dx = (\tan{4x})^3 \arctan{(\tan{4x})} + C$$ So, the final solution is $$\int \tan^{3}{4x} dx = \tan^3{4x} \arctan{(\tan{4x})} + C$$

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