Question

integral (tan^2x sec^2x)/(1-tan^6x)dx​

Answer 1

Answer for the Above Questions

Step by Step Explanation ☝️

Answer 2

$\bf \pink{\int \frac{ {tan}^{2}x \: {sec}^{2}x }{1 - {tan}^{6}x } dx = \frac{1}{6}log \bigg| \frac{1 + {tan}^{3}x }{1 - {tan}^{3}x} \bigg | + C}$

Given:

• $\int \frac{ {tan}^{2}x \: {sec}^{2}x }{1 - {tan}^{6}x } dx$

To find:

• Integrate the function.

Solution:

Concept/formula to be used:

• Apply substitution method.
• $\int \: \frac{1}{ {a}^{2} - {x}^{2} } dx = \frac{1}{2a} log \bigg| \frac{a + x}{a - x} \bigg | + C$

Step 1:

Put

${tan}^{3} x = t$

Differentiate with respect to x both sides.

$3 {tan}^{2} x. {sec}^{2} x = \frac{dt}{dx}$

$\because \frac{d}{dx}x^n=nx^{n-1}$

and

$\because \frac{d}{dx}tan\:x={sec}^{2} x$

or

${tan}^{2} x. {sec}^{2} x \: dx= \frac{1}{3} dt$

Substitute into function.

$\int \frac{ {tan}^{2}x \: {sec}^{2}x }{1 - {tan}^{6}x } dx =\frac{1}{3}\int \frac{ 1 }{1 - {t}^{2}}dt$

Step 2:

Integrate the function.

$\frac{1}{3}\int \frac{ 1 }{1 - {t}^{2}}dt = \frac{1}{3}. \frac{1}{2 \times 1} log \bigg| \frac{1 - t}{1 + t} \bigg | + C$

or

$\frac{1}{3}\int \frac{ 1 }{1 - {t}^{2}}dt = \frac{1}{6}log \bigg| \frac{1 + t}{1 - t} \bigg | + C$

Step 3:

Redo substitution.

Put

$t = {tan}^{3} x$

So,

$\int \frac{ {tan}^{2}x \: {sec}^{2}x }{1 - {tan}^{6}x } dx = \frac{1}{6}log \bigg| \frac{1 + {tan}^{3}x }{1 - {tan}^{3}x} \bigg | + C$

Thus,

$\bf \int \frac{ {tan}^{2}x \: {sec}^{2}x }{1 - {tan}^{6}x } dx = \frac{1}{6}log \bigg| \frac{1 + {tan}^{3}x }{1 - {tan}^{3}x} \bigg | + C$

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