The question
(a) Find : ∫ x + sin x 1 + cos x d x \displaystyle\int\frac{x + \sin x}{1 + \cos x}\,dx ∫ 1 + cos x x + sin x d x
OR
(b) Evaluate : ∫ 0 π 4 d x cos 3 x 2 sin 2 x \displaystyle\int_0^{\frac{\pi}{4}}\frac{dx}{\cos^3x\sqrt{2\sin 2x}} ∫ 0 4 π cos 3 x 2 sin 2 x d x
Exam answer Show all explanations
(a)
Let I = ∫ x + sin x 1 + cos x d x I = \displaystyle\int\frac{x + \sin x}{1 + \cos x}\,dx I = ∫ 1 + cos x x + sin x d x
Using 1 + cos x = 2 cos 2 x 2 1 + \cos x = 2\cos^2\frac{x}{2} 1 + cos x = 2 cos 2 2 x sin x = 2 sin x 2 cos x 2 \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} sin x = 2 sin 2 x cos 2 x
I = ∫ x + 2 sin x 2 cos x 2 2 cos 2 x 2 d x = ∫ x ( 1 2 sec 2 x 2 ) d x + ∫ tan x 2 d x \begin{aligned} I &= \int\frac{x + 2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}\,dx \\ &= \int x\left(\frac12\sec^2\frac{x}{2}\right)dx \\ &\quad + \int\tan\frac{x}{2}\,dx \end{aligned} I = ∫ 2 cos 2 2 x x + 2 sin 2 x cos 2 x d x = ∫ x ( 2 1 sec 2 2 x ) d x + ∫ tan 2 x d x Why?
The half-angle identities turn the denominator into a single square, 2 cos 2 x 2 2\cos^2\frac{x}{2} 2 cos 2 2 x
x 2 cos 2 x 2 = x ⋅ 1 2 sec 2 x 2 \frac{x}{2\cos^2\frac{x}{2}} = x\cdot\frac12\sec^2\frac{x}{2} 2 cos 2 2 x x = x ⋅ 2 1 sec 2 2 x 2 sin x 2 cos x 2 2 cos 2 x 2 = tan x 2 \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}} = \tan\frac{x}{2} 2 cos 2 2 x 2 sin 2 x cos 2 x = tan 2 x In detail: for the first part, write 1 cos 2 = sec 2 \frac{1}{\cos^2} = \sec^2 c o s 2 1 = sec 2
x 2 cos 2 x 2 = x ⋅ 1 2 ⋅ 1 cos 2 x 2 = x ⋅ 1 2 sec 2 x 2 \begin{aligned} \frac{x}{2\cos^2\frac{x}{2}} &= x\cdot\frac12\cdot\frac{1}{\cos^2\frac{x}{2}} \\ &= x\cdot\frac12\sec^2\frac{x}{2} \end{aligned} 2 cos 2 2 x x = x ⋅ 2 1 ⋅ cos 2 2 x 1 = x ⋅ 2 1 sec 2 2 x For the second part, cancel the 2 and one cos x 2 \cos\frac{x}{2} cos 2 x sin cos = tan \frac{\sin}{\cos} = \tan c o s s i n = tan
2 sin x 2 cos x 2 2 cos x 2 cos x 2 = sin x 2 cos x 2 = tan x 2 \begin{aligned} \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos\frac{x}{2}\cos\frac{x}{2}} &= \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \\ &= \tan\frac{x}{2} \end{aligned} 2 cos 2 x cos 2 x 2 sin 2 x cos 2 x = cos 2 x sin 2 x = tan 2 x Integrating the first integral by parts, with x x x ∫ 1 2 sec 2 x 2 d x = tan x 2 \int\frac12\sec^2\frac{x}{2}\,dx = \tan\frac{x}{2} ∫ 2 1 sec 2 2 x d x = tan 2 x
Why?
The first integral is a product of x x x
∫ u v d x = u ∫ v d x − ∫ ( d u d x ∫ v d x ) d x \begin{aligned} \int u\,v\,dx &= u\int v\,dx \\ &\quad - \int\left(\frac{du}{dx}\int v\,dx\right)dx \end{aligned} ∫ u v d x = u ∫ v d x − ∫ ( d x d u ∫ v d x ) d x Take u = x u = x u = x d u d x = 1 \frac{du}{dx} = 1 d x d u = 1 v = 1 2 sec 2 x 2 v = \frac12\sec^2\frac{x}{2} v = 2 1 sec 2 2 x
d d x tan x 2 = 1 2 sec 2 x 2 \frac{d}{dx}\tan\frac{x}{2} = \frac12\sec^2\frac{x}{2} d x d tan 2 x = 2 1 sec 2 2 x so ∫ v d x = tan x 2 \int v\,dx = \tan\frac{x}{2} ∫ v d x = tan 2 x
I = x tan x 2 − ∫ tan x 2 d x + ∫ tan x 2 d x \begin{aligned} I &= x\tan\frac{x}{2} - \int\tan\frac{x}{2}\,dx \\ &\quad + \int\tan\frac{x}{2}\,dx \end{aligned} I = x tan 2 x − ∫ tan 2 x d x + ∫ tan 2 x d x Why?
∫ x ( 1 2 sec 2 x 2 ) d x = x tan x 2 − ∫ ( 1 ) tan x 2 d x \begin{aligned} &\int x\left(\frac12\sec^2\frac{x}{2}\right)dx \\ &= x\tan\frac{x}{2} - \int (1)\tan\frac{x}{2}\,dx \end{aligned} ∫ x ( 2 1 sec 2 2 x ) d x = x tan 2 x − ∫ ( 1 ) tan 2 x d x By parts gives x tan x 2 − ∫ 1 ⋅ tan x 2 d x x\tan\frac{x}{2} - \int 1\cdot\tan\frac{x}{2}\,dx x tan 2 x − ∫ 1 ⋅ tan 2 x d x
Watch out: do not integrate tan x 2 \tan\frac{x}{2} tan 2 x C C C
∴ I = x tan x 2 + C \therefore\ I = x\tan\frac{x}{2} + C ∴ I = x tan 2 x + C
(b)
Let I = ∫ 0 π 4 d x cos 3 x 2 sin 2 x I = \displaystyle\int_0^{\frac{\pi}{4}}\frac{dx}{\cos^3x\sqrt{2\sin 2x}} I = ∫ 0 4 π cos 3 x 2 sin 2 x d x
Here 2 sin 2 x = 4 sin x cos x 2\sin 2x = 4\sin x\cos x 2 sin 2 x = 4 sin x cos x = 4 cos 2 x tan x = 4\cos^2x\tan x = 4 cos 2 x tan x 2 sin 2 x = 2 cos x tan x \sqrt{2\sin 2x} = 2\cos x\sqrt{\tan x} 2 sin 2 x = 2 cos x tan x
Why?
Use sin 2 x = 2 sin x cos x \sin 2x = 2\sin x\cos x sin 2 x = 2 sin x cos x sin x = cos x tan x \sin x = \cos x\tan x sin x = cos x tan x
2 sin 2 x = 2 ( 2 sin x cos x ) = 4 sin x cos x = 4 ( cos x tan x ) cos x = 4 cos 2 x tan x 4 cos 2 x tan x = 4 cos 2 x tan x = 2 cos x tan x \begin{aligned} 2\sin 2x &= 2(2\sin x\cos x) \\ &= 4\sin x\cos x \\ &= 4(\cos x\tan x)\cos x \\ &= 4\cos^2x\tan x \\[4pt] \sqrt{4\cos^2x\tan x} &= \sqrt4\,\sqrt{\cos^2x}\,\sqrt{\tan x} \\ &= 2\cos x\sqrt{\tan x} \end{aligned} 2 sin 2 x 4 cos 2 x tan x = 2 ( 2 sin x cos x ) = 4 sin x cos x = 4 ( cos x tan x ) cos x = 4 cos 2 x tan x = 4 cos 2 x tan x = 2 cos x tan x On [ 0 , π 4 ] \left[0, \frac{\pi}{4}\right] [ 0 , 4 π ] cos x > 0 \cos x > 0 cos x > 0 4 cos 2 x = 2 cos x \sqrt{4\cos^2x} = 2\cos x 4 cos 2 x = 2 cos x
The aim is to write everything in tan x \tan x tan x sec 2 x \sec^2x sec 2 x d d x tan x = sec 2 x \frac{d}{dx}\tan x = \sec^2x d x d tan x = sec 2 x
I = 1 2 ∫ 0 π 4 d x cos 4 x tan x = 1 2 ∫ 0 π 4 sec 2 x ⋅ sec 2 x tan x d x \begin{aligned} I &= \frac12\int_0^{\frac{\pi}{4}}\frac{dx}{\cos^4x\sqrt{\tan x}} \\ &= \frac12\int_0^{\frac{\pi}{4}}\frac{\sec^2x\cdot\sec^2x}{\sqrt{\tan x}}\,dx \end{aligned} I = 2 1 ∫ 0 4 π cos 4 x tan x d x = 2 1 ∫ 0 4 π tan x sec 2 x ⋅ sec 2 x d x Why?
The denominator is
cos 3 x ⋅ 2 cos x tan x = 2 cos 4 x tan x \begin{aligned} &\cos^3x\cdot2\cos x\sqrt{\tan x} \\ &= 2\cos^4x\sqrt{\tan x} \end{aligned} cos 3 x ⋅ 2 cos x tan x = 2 cos 4 x tan x and the 2 2 2 1 2 \frac12 2 1
Then
1 cos 4 x = sec 4 x = sec 2 x ⋅ sec 2 x \begin{aligned} \frac{1}{\cos^4x} &= \sec^4x \\ &= \sec^2x\cdot\sec^2x \end{aligned} cos 4 x 1 = sec 4 x = sec 2 x ⋅ sec 2 x One sec 2 x \sec^2x sec 2 x d x dx d x
Put tan x = t \tan x = t tan x = t sec 2 x d x = d t \sec^2x\,dx = dt sec 2 x d x = d t sec 2 x = 1 + tan 2 x = 1 + t 2 \sec^2x = 1 + \tan^2x = 1 + t^2 sec 2 x = 1 + tan 2 x = 1 + t 2
Why?
Differentiating tan x = t \tan x = t tan x = t sec 2 x d x = d t \sec^2x\,dx = dt sec 2 x d x = d t sec 2 x \sec^2x sec 2 x t t t sec 2 x = 1 + tan 2 x \sec^2x = 1 + \tan^2x sec 2 x = 1 + tan 2 x
When x = 0 x = 0 x = 0 t = 0 t = 0 t = 0 x = π 4 x = \frac{\pi}{4} x = 4 π t = 1 t = 1 t = 1
Why?
With a substitution in a definite integral, change the limits too: tan 0 = 0 \tan0 = 0 tan 0 = 0 tan π 4 = 1 \tan\frac{\pi}{4} = 1 tan 4 π = 1 x x x
I = 1 2 ∫ 0 1 1 + t 2 t d t = 1 2 ∫ 0 1 ( t − 1 2 + t 3 2 ) d t = 1 2 [ 2 t 1 2 + 2 5 t 5 2 ] 0 1 = 1 2 ( 2 + 2 5 ) − 0 \begin{aligned} I &= \frac12\int_0^1\frac{1 + t^2}{\sqrt t}\,dt \\ &= \frac12\int_0^1\left(t^{-\frac12} + t^{\frac32}\right)dt \\ &= \frac12\left[2t^{\frac12} + \frac25t^{\frac52}\right]_0^1 \\ &= \frac12\left(2 + \frac25\right) - 0 \end{aligned} I = 2 1 ∫ 0 1 t 1 + t 2 d t = 2 1 ∫ 0 1 ( t − 2 1 + t 2 3 ) d t = 2 1 [ 2 t 2 1 + 5 2 t 2 5 ] 0 1 = 2 1 ( 2 + 5 2 ) − 0 Why?
Divide each term by t = t 1 2 \sqrt t = t^{\frac12} t = t 2 1
1 t 1 / 2 = t − 1 2 , t 2 t 1 / 2 = t 3 2 \frac{1}{t^{1/2}} = t^{-\frac12},\quad \frac{t^2}{t^{1/2}} = t^{\frac32} t 1/2 1 = t − 2 1 , t 1/2 t 2 = t 2 3 Then use the power rule for integrals:
∫ t n d t = t n + 1 n + 1 \int t^n\,dt = \frac{t^{n + 1}}{n + 1} ∫ t n d t = n + 1 t n + 1 t 1 / 2 1 / 2 = 2 t 1 2 t 5 / 2 5 / 2 = 2 5 t 5 2 \begin{aligned} \frac{t^{1/2}}{1/2} &= 2t^{\frac12} \\ \frac{t^{5/2}}{5/2} &= \frac25t^{\frac52} \end{aligned} 1/2 t 1/2 5/2 t 5/2 = 2 t 2 1 = 5 2 t 2 5 At t = 1 t = 1 t = 1 2 + 2 5 = 12 5 2 + \frac25 = \frac{12}{5} 2 + 5 2 = 5 12 t = 0 t = 0 t = 0 0 0 0 1 2 ⋅ 12 5 = 6 5 \frac12\cdot\frac{12}{5} = \frac65 2 1 ⋅ 5 12 = 5 6
t = 1 : 2 ( 1 ) 1 2 + 2 5 ( 1 ) 5 2 = 2 + 2 5 t = 0 : 2 ( 0 ) + 2 5 ( 0 ) = 0 2 + 2 5 = 10 5 + 2 5 = 12 5 1 2 ⋅ 12 5 = 12 10 = 12 ÷ 2 10 ÷ 2 = 6 5 \begin{aligned} t = 1&:\ 2(1)^{\frac12} + \frac25(1)^{\frac52} = 2 + \frac25 \\ t = 0&:\ 2(0) + \frac25(0) = 0 \\[4pt] 2 + \frac25 &= \frac{10}{5} + \frac25 = \frac{12}{5} \\ \frac12\cdot\frac{12}{5} &= \frac{12}{10} \\ &= \frac{12 \div 2}{10 \div 2} = \frac65 \end{aligned} t = 1 t = 0 2 + 5 2 2 1 ⋅ 5 12 : 2 ( 1 ) 2 1 + 5 2 ( 1 ) 2 5 = 2 + 5 2 : 2 ( 0 ) + 5 2 ( 0 ) = 0 = 5 10 + 5 2 = 5 12 = 10 12 = 10 ÷ 2 12 ÷ 2 = 5 6 ∴ I = 6 5 \therefore\ I = \dfrac65 ∴ I = 5 6
Where this comes from Chapter 7, Integrals : Integration by parts; definite by substitution.
Question type: (a) Integration by parts ; (b) Substitution
How to spot it: click to see (a) Type 8. Integration by parts: a product of two different kinds
Spot it: a product such as x sin x x\sin x x sin x x e x xe^x x e x x log x x\log x x log x log x \log x log x tan − 1 x \tan^{-1}x tan − 1 x
Here: The half-angle forms split the integrand into x ⋅ 1 2 sec 2 x 2 x\cdot\frac12\sec^2\frac{x}{2} x ⋅ 2 1 sec 2 2 x tan x 2 \tan\frac{x}{2} tan 2 x u = x u = x u = x − ∫ tan x 2 d x -\int\tan\frac{x}{2}\,dx − ∫ tan 2 x d x
(b) Type 2. Substitution: when the derivative is hiding in the question
Spot it: a function and its derivative both appear, like log x \log x log x 1 x \frac{1}{x} x 1 x 2 + 1 x^2 + 1 x 2 + 1 x x x
Here: With 2 sin 2 x = 2 cos x tan x \sqrt{2\sin 2x} = 2\cos x\sqrt{\tan x} 2 sin 2 x = 2 cos x tan x sec 2 x ⋅ sec 2 x 2 tan x \frac{\sec^2x\cdot\sec^2x}{2\sqrt{\tan x}} 2 t a n x s e c 2 x ⋅ s e c 2 x tan x \tan x tan x sec 2 x \sec^2x sec 2 x tan x = t \tan x = t tan x = t sec 2 x \sec^2x sec 2 x 1 + t 2 1 + t^2 1 + t 2
Quick lesson · Worked NCERT solutions · Interactive lesson