Find the integrals of the functions- -sin 32 x-1 d x

∫ sin^3 (2x+1)dx=∫ sin^2 (2x+1).sin (2x+1)dx =∫ [1 cos^2(2x+1)]sin(2x+1)dx (∴ sin^2 x =1 cos^2 x) Let cos(2x+1)=t ⇒ 2 sin (2x+1)dx =dt ⇒ sin (2x+1)dx = d ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

Find the inte...

Question

Find the integrals of the functions.
$\int s i n^{3} (2 x + 1) d x$

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Solution

$\int s i n^{3} (2 x + 1) d x = \int s i n^{2} (2 x + 1) . s i n (2 x + 1) d x = \int [1 - c o s^{2} (2 x + 1)] s i n (2 x + 1) d x (∴ s i n^{2} x = 1 - c o s^{2} x)$
Let $c o s (2 x + 1) = t \Rightarrow - 2 s i n (2 x + 1) d x = d t \Rightarrow s i n (2 x + 1) d x = - \frac{d t}{2}$
$∴ \int s i n^{3} (2 x + 1) d x = \frac{- 1}{2} \int (1 - t^{2}) d t = \frac{- 1}{2} (t - \frac{t^{3}}{3}) + C = \frac{- 1}{2} (c o s (2 x + 1) - \frac{c o s^{3} (2 x + 1)}{3}) + C = \frac{- c o s (2 x + 1)}{2} + \frac{c o s^{3} (2 x + 1)}{6} + C$

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Find the integrals of the functions. ∫sin3(2x+1)dx

∫sin3(2x+1)dx=∫sin2(2x+1).sin(2x+1)dx=∫[1−cos2(2x+1)]sin(2x+1)dx (∴sin2x=1−cos2x) Let cos(2x+1)=t⇒−2sin(2x+1)dx=dt⇒sin(2x+1)dx=−dt2 ∴∫sin3(2x+1)dx=−12∫(1−t2)dt=−12(t−t33)+C=−12(cos(2x+1)−cos3(2x+1)3)+C=−cos(2x+1)2+cos3(2x+1)6+C

Find the integrals of the functions.
$\int s i n^{3} (2 x + 1) d x$