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Let $n\geq 2$ be a natural numberr and $0<\Theta <\pi/2$

Then $\int \frac{(sin^{n}\Theta -sin\Theta )^{1/n}\cos\Theta }{\sin^{n+1}\Theta}d\Theta$ is equal to: (where C is a constant of integration)
Option 1)
$\frac{n }{n^{2}-1}(1-\frac{1}{sin^{n-1}\theta})^{n+1/n}+C$

Option 2)
$\frac{n}{n^{2}+1}(1-\frac{1 }{\sin^{n-1}\Theta})^{n+1/n} +C$

Option 3)

$n/n^{2}-1(1+1/sin^{n-1}\Theta )^{n+1/n}+c$
Option 4)

$\frac{n}{n^{2}-1}(1-\frac{1}{\sin^{n+1}\Theta })^{n+1/n}+C$

Answers (1)

best_answer

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

- wherein

Since $\int f(x)dx=\int f(t)dt=\int f(\theta )d\theta$ all variables must be converted into single variable , $\left ( t\, or\ \theta \right )$

Indefinite integrals for Algebraic functions -

$\frac{\mathrm{d}}{\mathrm{d} x} \frac{\left ( x^{n+1} \right )}{n+1}=x^{n}$ so $\int x^{n}dx=\frac{x^{n+1}}{n+1}$

- wherein

Where $n\neq-1$

$\int \frac{(\sin ^{n}\theta -\sin \theta )^{\frac{1}{n}}\cos \theta }{\sin ^{n+1}\theta }d\theta$

This can be written as

$\int \frac{\sin \theta (1-\frac{1}{\sin ^{n-1}\theta })^{\frac{1}{n}}}{\sin ^{n+1}\theta \sec \theta }d\theta$

put $1-\frac{1}{\sin ^{n-1}\theta }=t$

$\frac{n-1}{\sin ^{n}\theta }\cos \theta \: \: d\theta =dt$

Now,

$\frac{1}{n-1}\int (t)^{}\frac{1}{n}dt=\frac{1}{n-1}\frac{(t)^{\frac{1}{n}+1}}{\frac{1}{n}+1}+C$

$=\frac{1}{n-1}\frac{t^{(\frac{1}{n}+1)}}{\frac{1}{n}+1}+C$

$=\frac{1}{n-1}(1-\frac{1}{\sin ^{n-1}\theta })^{\frac{1}{n}+1}+C$

Option 1)
$\frac{n }{n^{2}-1}(1-\frac{1}{sin^{n-1}\theta})^{n+1/n}+C$

Option 2)

$\frac{n}{n^{2}+1}(1-\frac{1 }{\sin^{n-1}\Theta})^{n+1/n} +C$

Option 3)

$n/n^{2}-1(1+1/sin^{n-1}\Theta )^{n+1/n}+c$

Option 4)

$\frac{n}{n^{2}-1}(1-\frac{1}{\sin^{n+1}\Theta })^{n+1/n}+C$

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