The integral of $\frac{{cos}^{3} x}{{sin}^{2} x + sin x}$ gives $f (x)$ with const . of integration as zero. find $f (π / 2)$

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Solution

$I = \int \frac{(1 - {sin}^{2} x) cos x}{sin x (1 + sin x)} d x$

$= \int \frac{1 - sin x}{sin x} cos x d x$

Put $t = sin x$
$\Rightarrow cos x d x = d t$

So, $I = \int (\frac{1}{t} - 1) d t$

$= log t - t + C$
$\Rightarrow I = log sin x - sin x + C$
$\Rightarrow I = log sin x - sin x$ (Given $C = 0$ )
Therefore, $f (\frac{π}{2}) = - 1$

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, then the value of

f (π)

= if the constant of integration is equal to zero.

Gives f(x) with constant of integration zero then find f(1)

\int \frac{log x}{{(1 + log x)}^{2}} d x

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The integral of cos3xsin2x+sinx gives f(x) with const . of integration as zero. find f(π/2)

I=∫(1−sin2x)cosxsinx(1+sinx)dx=∫1−sinxsinxcosxdx Put t=sinx ⇒cosxdx=dtSo, I=∫(1t−1)dt =logt−t+C⇒I=logsinx−sinx+C ⇒I=logsinx−sinx (Given C=0)Therefore, f(π2)=−1

The integral of $\frac{{cos}^{3} x}{{sin}^{2} x + sin x}$ gives $f (x)$ with const . of integration as zero. find $f (π / 2)$