Can we use integration by substitution on these integrals?
Question
By using the substitution u = 1 − x 2 u = 1 - x^2 u = 1 − x 2
∫ 0 1 2 x ( 1 − x 2 ) 2 d x . \begin{equation*} \int_0^{\frac{1}{2}} \frac{x}{(1-x^2)^2} \:dx. \end{equation*} ∫ 0 2 1 ( 1 − x 2 ) 2 x d x .
Evaluate
∫ 0 π / 6 sin θ d θ cos 3 θ . \begin{equation*} \int_0^{\pi/6} \frac{\sin \theta \:d\theta}{\cos^3 \theta}. \end{equation*} ∫ 0 π /6 cos 3 θ sin θ d θ . Solution
If we take
u = 1 − x 2 u = 1 - x^2 u = 1 − x 2
, then
d u d x = − 2 x ⟹ d u = − 2 x d x \begin{equation*} \frac{du}{dx} = -2x \implies du = -2x \:dx \end{equation*} d x d u = − 2 x ⟹ d u = − 2 x d x and therefore, by substitution,
∫ 0 1 2 x ( 1 − x 2 ) 2 d x = ∫ 1 − 0 1 − ( 1 2 ) 2 d u − 2 u 2 = 1 2 ∫ 3 4 1 d u u 2 = 1 2 [ u − 1 − 1 ] 3 4 1 = − 1 2 ( 1 − 4 3 ) = 1 6 . \begin{align*} \int_0^{\frac{1}{2}} \frac{x}{(1-x^2)^2} \:dx = \int_{1-0}^{1-\left(\frac{1}{2}\right)^2} \frac{du}{-2u^2} &= \frac{1}{2} \int_{\frac{3}{4}}^1 \frac{du}{u^2} \\ &= \frac{1}{2} \left[ \frac{u^{-1}}{-1} \right]_{\frac{3}{4}}^1 \\ &= -\frac{1}{2} \left( 1 - \frac{4}{3} \right) \\ &= \frac{1}{6}. \end{align*} ∫ 0 2 1 ( 1 − x 2 ) 2 x d x = ∫ 1 − 0 1 − ( 2 1 ) 2 − 2 u 2 d u = 2 1 ∫ 4 3 1 u 2 d u = 2 1 [ − 1 u − 1 ] 4 3 1 = − 2 1 ( 1 − 3 4 ) = 6 1 . Note how we can make an integral look simpler by factoring out constants, including negatives which we can take care of by swapping the limits over.
In part (i) the x 2 x^2 x 2 x x x sin θ \sin \theta sin θ cos θ \cos \theta cos θ
Take
u = cos θ u = \cos \theta u = cos θ
, so that
d u d θ = − sin θ ⟹ d u = − sin θ d θ . \begin{equation*} \frac{du}{d\theta} = -\sin \theta \implies du = - \sin \theta \:d\theta. \end{equation*} d θ d u = − sin θ ⟹ d u = − sin θ d θ . By substitution, then,
∫ 0 π / 6 sin θ d θ cos 3 θ = ∫ cos 0 cos π 6 − d u u 3 = − [ u − 2 − 2 ] 1 3 / 2 = 1 2 ( ( 2 3 ) 2 − 1 ) = 1 6 . \begin{align*} \int_0^{\pi/6} \frac{\sin \theta \:d\theta}{\cos^3 \theta} = \int_{\cos 0}^{\cos \frac{\pi}{6}} \frac{-du}{u^3} &= - \left[ \frac{u^{-2}}{-2} \right]_1^{\sqrt{3}/2} \\ &= \frac{1}{2} \left( \left( \frac{2}{\sqrt{3}} \right)^2 - 1 \right) \\ &= \frac{1}{6}. \end{align*} ∫ 0 π /6 cos 3 θ sin θ d θ = ∫ c o s 0 c o s 6 π u 3 − d u = − [ − 2 u − 2 ] 1 3 /2 = 2 1 ( ( 3 2 ) 2 − 1 ) = 6 1 .