If ∫ 1 + c o s x e x ( 1 + s i n x ) dx = e x f ( x ) + C , then f ( x ) is equal to

∫ e x ( 1 + c o s x ) ( 1 + s i n x ) d x = ∫ e x [ 2 1 sec 2 2 x + tan 2 x ] d x = 2 1 ∫ e x sec 2 2 x d x + ∫ e x tan 2 x d x = e x tan 2 x + C But I = e x f ( x ) + C (given) ∴ f ( x ) = tan 2 x

Mathematics - Indefinite Integration Question with Solution | TestHub

MathematicsIndefinite IntegrationBy partEasy2 minPYQ_2023

MathematicsEasysingle choice

If $\int \frac{e ^{x} ( 1 + s i n x ) dx}{1 + c o s x} = e^{x} f (x) + C$ , then $f (x)$ is equal to

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Answer:

Solution:

$\int e^{x} \frac{( 1 + s i n x )}{( 1 + c o s x )} d x$
$= \int e^{x} [\frac{1}{2} sec^{2} \frac{x}{2} + tan \frac{x}{2}] d x$
$= \frac{1}{2} \int e^{x} sec^{2} \frac{x}{2} d x + \int e^{x} tan \frac{x}{2} d x$
$= e^{x} tan \frac{x}{2} + C$
But $I = e^{x} f (x) + C$ (given)
$∴ f (x) = tan \frac{x}{2}$

Stream:BITSATSubject:MathematicsTopic:Indefinite IntegrationSubtopic:By part

⏱ 2mℹ️ Source: PYQ_2023

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