Question

$\displaystyle\sf\int\dfrac{tan^3\:(ln\:x)\:dx}{x}$

Appropriate answers needed ^^"​

Answer 1

$\sf{\displaystyle\int \frac{\tan ^{3}(\ln x) d x}{x}}$

• Remove parentheses

$\to\sf{\displaystyle\int \frac{\tan ^{3}\ln x\:dx}{x}}$

• Use Integration by Substitution

Let $\sf{u=In x}$ , $\sf{du=\dfrac{1}{x}dx}$

• Using u and du above, rewrite $\sf{\displaystyle\int \frac{\tan ^{3}\ln x\:dx}{x}}$

$\to\sf{\int \tan ^{3} u d u}$

• Use Pythagorean Identities $\sf{\tan^{2} u=\sec^{2} u-1}$

$\to\sf{\int\left(\sec ^{2} u-1\right) \tan u d u}$

$\to\sf{\int \tan u \sec ^{2} u-\tan u d u}$

• Use Sum Rule : $\sf{\int f(x)+g(x) \dx=\int f(x) \dx+\int g(x) \, dx}$

$\to\sf{\int \tan u \sec ^{2} u d u-\int \tan u d u}$

• Use Integration by Substitution on $\sf{\int \tan{u}\sec^{2}u \du}$

Let $\sf{w=\tan u}$ , $\sf{dw=sec^2udu}$

• Using w and dw above, rewrite $\sf{\int \tan{u}\sec^{2}u \du}$

$\to\sf{\int w d w}$

• Use Power Rule $\sf{\displaystyle\int x^{n} d x=\frac{x^{n+1}}{n+1}+C .}$

$\to\sf{\dfrac{w^2}{2}}$

• Substitute $\sf{w=\tan x}$ back into the original integral.

$\to\sf{\dfrac{\tan ^{2} u}{2}}$

• Rewrite the integral with the completed substitution.

$\to\sf{\displaystyle\frac{\tan ^{2} u}{2}-\int \tan u d u}$

• Use Trigonometric Integration : The integral of $\sf{\tan{u}}$ is $\sf{\ln{(\sec{u})}}$

$\to\sf{\dfrac{\tan ^{2} u}{2}-\ln (\sec u)}$

• Substitute $\sf{u=lnx}$ back into the original integral.

$\to\sf{\displaystyle\frac{\tan ^{2} \ln x}{2}-\ln (\sec (\ln x))}$

• Add a Constant

$\boxed{\to\sf{\displaystyle\frac{\tan ^{2} \ln x}{2}-\ln (\sec (\ln x))+C}}$

Answer 2

Answer:

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