Integrate sin-e-cos

Calculus Level 4

∫ 0 π / 2 e sin ⁡ x cos ⁡ 3 x d x \large \displaystyle \int_{0}^{\pi /2} e^{\sin x}\cos^3 x \ dx

Find the closed form of the integral above. Give your answer to 2 decimal places.


The answer is 1.

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Harsh Shrivastava by Harsh Shrivastava , 20, India

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2 solutions

Let sin ⁡ x = y \sin x=y , then

∫ 0 π / 2 e sin ⁡ x ( 1 − cos ⁡ 2 x ) d x = ∫ 0 1 e y ( 1 − y 2 ) d y \int_{0}^{\pi/2} e^{\sin x}(1-\cos^2 x) dx = \int_{0}^{1} e^y (1-y^2) dy

e y ( 1 − y 2 ) = − e y ( y 2 − 1 ) = − e y ( y 2 − 2 y + 1 + 2 ( y − 1 ) ) = − e y ( y − 1 ) 2 − 2 e y ( y − 1 ) e^y(1-y^2) = -e^y(y^2-1) = -e^y(y^2-2y+1+2(y-1)) = -e^y(y-1)^2 - 2e^y(y-1)

− e y ( y − 1 ) 2 − 2 e y ( y − 1 ) = − ( y − 1 ) 2 d d y e y + e y d d y ( y − 1 ) 2 = d d y e y ( 1 − y 2 ) -e^y(y-1)^2 - 2e^y(y-1) = -(y-1)^2\frac{d}{dy} e^y+e^y\frac{d}{dy}(y-1)^2=\frac{d}{dy}e^y(1-y^2)

Hence,

∫ 0 1 e y ( 1 − y 2 ) d y = e y ( 1 − y 2 ) ∣ 0 1 = 1 \int_{0}^{1} e^y (1-y^2) dy = \left.e^y(1-y^2) \right|_{0}^{1}=\boxed{1}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Same way. Try solving this

Samara Simha Reddy - 5 years, 1 month ago

Substituting e s i n x = t e^{sinx} =t gives ∫ ( 1 − ln ⁡ 2 t ) d t \int (1-\ln^{2} t) dt which can be evaluated using IBP.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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