Can we carry out these three tricky integrals?
Question
Find
∫ x 2 x − 1 d x \displaystyle \int x \sqrt{2x-1} \:dx ∫ x 2 x − 1 d x
∫ 1 2 x − x 2 d x \displaystyle \int \frac{1}{\sqrt{2x - x^2}} \:dx ∫ 2 x − x 2 1 d x
Evaluate ∫ 0 π / 4 sec 4 x d x \displaystyle \int_0^{\pi/4} \sec^4 x \:dx ∫ 0 π /4 sec 4 x d x
Suggestion
(a)(ii) Find ∫ 1 2 x − x 2 d x \displaystyle \int \frac{1}{\sqrt{2x - x^2}} \:dx ∫ 2 x − x 2 1 d x
Could we try completing the square on the denominator?
What do we get if we differentiate tan x \tan x tan x
Solution
Let
u = 2 x − 1 u = 2x - 1 u = 2 x − 1
, so that
d u = 2 d x du = 2 \: dx d u = 2 d x
. Then
∫ x 2 x − 1 d x = ∫ u + 1 2 u d u 2 = 1 4 ∫ u 3 / 2 + u 1 / 2 d u = 1 4 ( u 5 / 2 5 2 + u 3 / 2 3 2 + C ) = 1 10 u 5 / 2 + 1 6 u 3 / 2 + C = 1 10 ( 2 x − 1 ) 5 / 2 + 1 6 ( 2 x − 1 ) 3 / 2 + C , \begin{align*} \int x \sqrt{2x-1} \:dx = \int \frac{u+1}{2} \sqrt{u} \:\frac{du}{2} &= \frac{1}{4} \int u^{3/2} + u^{1/2} \:du \\ &= \frac{1}{4} \left( \frac{u^{5/2}}{\frac{5}{2}} + \frac{u^{3/2}}{\frac{3}{2}} + C \right) \\ &= \frac{1}{10} u^{5/2} + \frac{1}{6} u^{3/2} + C\\ &= \frac{1}{10} (2x-1)^{5/2} + \frac{1}{6} (2x-1)^{3/2} + C,\\ \end{align*} ∫ x 2 x − 1 d x = ∫ 2 u + 1 u 2 d u = 4 1 ∫ u 3/2 + u 1/2 d u = 4 1 ( 2 5 u 5/2 + 2 3 u 3/2 + C ) = 10 1 u 5/2 + 6 1 u 3/2 + C = 10 1 ( 2 x − 1 ) 5/2 + 6 1 ( 2 x − 1 ) 3/2 + C , where C C C
We can write
2 x − x 2 = − ( x 2 − 2 x ) = − ( ( x − 1 ) 2 − 1 ) = 1 − ( x − 1 ) 2 \begin{equation*} 2x - x^2 = -(x^2 - 2x) = -\left( (x-1)^2 - 1 \right) = 1 - (x-1)^2 \end{equation*} 2 x − x 2 = − ( x 2 − 2 x ) = − ( ( x − 1 ) 2 − 1 ) = 1 − ( x − 1 ) 2 so that
∫ 1 2 x − x 2 d x = ∫ 1 1 − ( x − 1 ) 2 d x . \begin{equation*} \int \frac{1}{\sqrt{2x - x^2}} \:dx = \int \frac{1}{\sqrt{1 - (x-1)^2}} \:dx. \end{equation*} ∫ 2 x − x 2 1 d x = ∫ 1 − ( x − 1 ) 2 1 d x . Take
sin θ = x − 1 \sin \theta = x - 1 sin θ = x − 1
, so that
cos θ d θ = d x \cos \theta \: d\theta = dx cos θ d θ = d x
and
∫ 1 1 − ( x − 1 ) 2 d x = ∫ cos θ 1 − sin 2 θ d θ = ∫ cos θ cos θ d θ = θ + C = sin − 1 ( x − 1 ) + C , \begin{align*} \int \frac{1}{\sqrt{1 - (x-1)^2}} \:dx = \int \frac{\cos \theta}{\sqrt{1 - \sin^2 \theta}} \:d\theta &= \int \frac{\cos \theta}{\cos \theta} \:d\theta \\ &= \theta + C \\ &= \sin^{-1} (x-1) + C, \end{align*} ∫ 1 − ( x − 1 ) 2 1 d x = ∫ 1 − sin 2 θ cos θ d θ = ∫ cos θ cos θ d θ = θ + C = sin − 1 ( x − 1 ) + C , where C C C
We have that
tan 2 x + 1 = sec 2 x . \begin{equation*} \tan^2 x + 1 = \sec^2 x. \end{equation*} tan 2 x + 1 = sec 2 x . By the standard result (which can be proved using the
quotient rule
), we have that
d d x tan x = sec 2 x . \begin{equation*} \frac{d}{dx} \tan x = \sec^2 x. \end{equation*} d x d tan x = sec 2 x . and therefore
∫ sec 4 x d x = ∫ sec 2 x sec 2 x d x = ∫ sec 2 x tan 2 x + sec 2 x d x . \begin{equation*} \int \sec^4 x \:dx = \int \sec^2 x \sec^2 x\:dx = \int \sec^2 x\tan^2 x + \sec^2 x \:dx. \end{equation*} ∫ sec 4 x d x = ∫ sec 2 x sec 2 x d x = ∫ sec 2 x tan 2 x + sec 2 x d x . The second term we can integrate by the standard result above. For the first term, we substitute u = tan x u=\tan x u = tan x ∫ sec 2 x tan 2 x d x = ∫ u 2 d u = tan 3 x 3 + C . \int \sec^2 x\tan^2 x\:dx = \int u^2\:du = \frac{\tan^3 x}{3}+C. ∫ sec 2 x tan 2 x d x = ∫ u 2 d u = 3 t a n 3 x + C . ∫ 0 π / 4 sec 4 x d x = [ tan 3 x 3 + tan x ] 0 π / 4 = 4 3 . \int_0^{\pi/4} \sec^4 x \:dx = \left[\frac{\tan^3 x}{3}+\tan x\right]_0^{\pi/4} = \frac{4}{3}. ∫ 0 π /4 sec 4 x d x = [ 3 t a n 3 x + tan x ] 0 π /4 = 3 4 .