Question

Evaluate the following ​step by step

Answer 1

Find the value of the integral cos^3x (lower limit = 0, upper limit = π/2)

Answer:

2/3

Step-by-step explanation:

$\implies \sf{\int\limits^{\pi/2}_0 {cos^3x} \, dx }$

$\implies \sf{\int\limits^{\pi/2}_0 {(cos^2x)cosx} \, dx }$

$\implies \sf{\int\limits^{\pi/2}_0 {(1-sin^2x)cosx} \, dx }$

$\sf{Let\;sinx\: = t,}\\\sf{ differentiate\:w.r.t.\:x}\\\sf{cosx.dx = dt},\;\;\;\; therefore$

$\\\implies \sf{\int\limits^{\pi/2}_0 {1} \, dt - \int\limits^{\pi/2}_0 {t^2} \, dt}$

$\\\\\implies \sf{t|_0 ^{\pi/2} - \bigg[\dfrac{t^3}{3}\bigg]\bigg|_0 ^{\pi/2} }$

$\\\\\sf{\implies [sin(\pi/2) - sin(0)] - \dfrac{1}{3}[sin^3(\pi/2) - sin^3(0)]}$

$\\\\\sf{\implies (1-0) - (1/3) [1 - 0]}$

$\\\\\implies 1 - 1/3 \\\\\implies 2/3$

Answer 2

Answer:

2/3

Step-by-step explanation:

- break cos³x as cos²x.cosx .

- convert cos²x in 1 - sin²x , you will get (1-sin²x).cosx .

- take t = sin x , so that it's derivative become cosx which is already given.

- now integrate 1-t² { as (1 - t²) = (1 - sin²x) and dt = cosx) }

- after integration, right the value of t as sinx in integral t - t³/3.

- now solve the definite integral sinx - (sinx)³/3 by putting the upper limit π/2 and lower limit 0.

- You will get 2/3. :)

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