Trigonometric integrals. Evaluate the following integrals. \int

Paganellash

Paganellash

Answered question

2021-12-03

Trigonometric integrals. Evaluate the following integrals.
0π(1cos2x)32dx

Answer & Explanation

Elizabeth Witte

Elizabeth Witte

Beginner2021-12-04Added 24 answers

Step 1
To evaluate the given integrals , 0π(1cos2x)32dx.
Solution:
Let us denote the given integral by I as,
I=0π(1cos2x)32dx
Since we know the identity as, 1cos2x=sin2x,
So the integral becomes,
I=0π(sin2x)32dx
So further solving the integral gives,
Step 2
I=0π(sin3x)dx
=0π(sin2x)sinxdx
=0π(1cos2x)(sinx)dx
Now using Substitution method, Put cosx=t, we get sinxdx=dt
Therefore, the given integral changes with upper limit, cosπ=1 and lower limit, cos0=1 as,
I=11(1t2)dt
I=(tt33)11
Step 3
Further, solving the integral gives,
I=(t33t)11
=(13+1)(131)
=23+23
=43
Hence, the value of the integral is 43.

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