Find - -sin x -logcos x d x

I=∫sin x logcos x dx Put cos x=t ⇒ nbsp; sin x dx=dt ⇒ I= ∫log t dt Using integration by parts, ∫ uv dx=u∫ v dx ∫du/dx (∫ v dx ) dx Here, u and v accordi ...

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Byju's Answer

Standard XII

Mathematics

Differentiation of Inverse Trigonometric Functions

Find : ∫sin x...

Question

$Find : \int sin x \cdot log cos x d x .$

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Solution

$I = \int sin x log cos x d x Put cos x = t \Rightarrow - sin x d x = d t \Rightarrow I = - \int log t d t$
Using integration by parts,
$\int u v d x = u \int v d x - \int \frac{d u}{d x} (\int v d x) d x$
$Here, u and v according to ILATE rule are log t and 1 respectively.$

$∴ I = - t {log}_{e} t + \int t \cdot \frac{1}{t} d t = - t {log}_{e} t + t + C$
$= t (1 - {log}_{e} t) + C = cos x (1 - {log}_{e} cos x) + C$

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Find:∫sin x⋅logcos x dx.

I=∫sin x logcos x dxPut cos x=t⇒−sin x dx=dt⇒I=−∫logt dt Using integration by parts, ∫uv dx=u∫v dx−∫dudx (∫v dx) dx Here, u and v according to ILATE rule are log t and 1 respectively. ∴I=−tloget+∫t⋅1t dt=−tloget+t+C =t(1−loget)+C=cos x(1−logecos x)+C

$Find : \int sin x \cdot log cos x d x .$