Get the answer to "Evaluate the following integrals. ∫sin2(θ+π6)dθ"

Maiclubk

Answered question

2021-10-08

Evaluate the following integrals.

\int \sin^{2} (θ + \frac{π}{6}) d θ

Answer & Explanation

Roosevelt Houghton

Skilled2021-10-09Added 106 answers

Given that:
$\int \sin^{2} (θ + \frac{π}{6}) d θ$
Solving the integral:
We have,
$\int \sin^{2} (θ + \frac{π}{6}) d θ$
Substitute,
$θ + \frac{π}{6} = t$
$\Rightarrow d θ = d t$
$\Rightarrow \int \sin^{2} (θ + \frac{π}{6}) d θ = \int \sin^{2} (t) d t$
Using the identity:
$\cos 2 t = 1 - 2 \sin^{2} t$
$\Rightarrow 2 \sin^{2} t = 1 - \cos 2 t$
$\Rightarrow \sin^{2} t = \frac{1 - \cos 2 t}{2}$
$\Rightarrow \int \sin^{2} (t) d t = \int \frac{1 - \cos 2 t}{2} d t$
$\Rightarrow \sin^{2} (t) d t = \int \frac{1}{2} d t - \frac{1}{2} \cos 2 t d t$
$\Rightarrow \sin^{2} (t) d t = \frac{t}{2} - \frac{\sin 2 t}{4} + C$
$\int \sin^{2} (θ + \frac{π}{6}) d θ = \frac{1}{2} [θ + \frac{π}{6} - \frac{1}{2} (\sin 2 (θ + \frac{π}{6}))] + C$

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