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CBSE Board Examination 2024 · Section D · 5 marks

Q33: Evaluate ∫0π/4sin⁡x+cos⁡x9+16sin⁡2x dx\int_0^{\pi/4}\frac{\sin x + \cos x}{9 + 16\sin 2x}\,dx⁠, or ∫0π/2sin⁡2xtan⁡−1(sin⁡x) dx\int_0^{\pi/2}\sin 2x\tan^{-1}(\sin x)\,dx

Board 2024Ch 7 · Integrals
The question

(a) Evaluate :

∫0π4sin⁡x+cos⁡x9+16sin⁡2x dx\int_0^{\frac{\pi}{4}}\frac{\sin x + \cos x}{9 + 16\sin 2x}\,dx

OR

(b) Evaluate :

∫0π2sin⁡2xtan⁡−1(sin⁡x) dx\int_0^{\frac{\pi}{2}}\sin 2x\tan^{-1}(\sin x)\,dx
Exam answer

(a)

Let I=∫0π/4sin⁡x+cos⁡x9+16sin⁡2x dxI = \displaystyle\int_0^{\pi/4}\frac{\sin x + \cos x}{9 + 16\sin 2x}\,dx⁠.

Put sin⁡x−cos⁡x=t\sin x - \cos x = t⁠, so that (cos⁡x+sin⁡x) dx=dt(\cos x + \sin x)\,dx = dt⁠.

Why?

The numerator sin⁡x+cos⁡x\sin x + \cos x is exactly the derivative of sin⁡x−cos⁡x\sin x - \cos x⁠:

ddx(sin⁡x−cos⁡x)=cos⁡x+sin⁡x\frac{d}{dx}(\sin x - \cos x) = \cos x + \sin x

So this choice of tt uses up the whole numerator with dxdx⁠. We only need sin⁡2x\sin 2x in terms of tt⁠, which the next line gives.

Squaring, sin⁡2x+cos⁡2x−sin⁡2x=t2\sin^2x + \cos^2x - \sin 2x = t^2⁠, so sin⁡2x=1−t2\sin 2x = 1 - t^2⁠.

Why?

Squaring sin⁡x−cos⁡x\sin x - \cos x gives sin⁡2x+cos⁡2x−2sin⁡xcos⁡x\sin^2x + \cos^2x - 2\sin x\cos x⁠. Then use

sin⁡2x+cos⁡2x=1\sin^2x + \cos^2x = 1 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \sin 2x t2=(sin⁡x−cos⁡x)2=sin⁡2x−2sin⁡xcos⁡x+cos⁡2x=(sin⁡2x+cos⁡2x)−2sin⁡xcos⁡x=1−sin⁡2xt2=1−sin⁡2xsin⁡2x=1−t2\begin{aligned} t^2 &= (\sin x - \cos x)^2 \\ &= \sin^2x - 2\sin x\cos x \\ &\quad + \cos^2x \\ &= (\sin^2x + \cos^2x) \\ &\quad - 2\sin x\cos x \\ &= 1 - \sin 2x \\[4pt] t^2 &= 1 - \sin 2x \\ \sin 2x &= 1 - t^2 \end{aligned}

(The last line adds sin⁡2x\sin 2x to both sides and subtracts t2t^2⁠.)

When x=0x = 0⁠, t=−1t = -1⁠; when x=π4x = \frac{\pi}{4}⁠, t=0t = 0⁠.

Why?

At x=0x = 0⁠: t=sin⁡0−cos⁡0=0−1=−1t = \sin 0 - \cos 0 = 0 - 1 = -1⁠.

At x=π4x = \frac{\pi}{4}⁠: sin⁡π4=cos⁡π4=12\sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{1}{\sqrt2}⁠, so t=0t = 0⁠.

t=12−12=0t = \frac{1}{\sqrt2} - \frac{1}{\sqrt2} = 0
I=∫−10dt9+16(1−t2)=∫−10dt25−16t2=116∫−10dt(54)2−t2\begin{aligned} I &= \int_{-1}^0\frac{dt}{9 + 16(1 - t^2)} \\ &= \int_{-1}^0\frac{dt}{25 - 16t^2} \\ &= \frac{1}{16}\int_{-1}^0\frac{dt}{\left(\frac54\right)^2 - t^2} \end{aligned}
Why?

First 9+16−16t2=25−16t29 + 16 - 16t^2 = 25 - 16t^2⁠.

9+16(1−t2)=9+16−16t2=25−16t2\begin{aligned} 9 + 16(1 - t^2) &= 9 + 16 - 16t^2 \\ &= 25 - 16t^2 \end{aligned}

To reach a standard form, the coefficient of t2t^2 must be 1, so take out 16:

25−16t2=16(2516−t2)25 - 16t^2 = 16\left(\frac{25}{16} - t^2\right)

and 2516=(54)2\frac{25}{16} = \left(\frac54\right)^2⁠.

(Check: 16×2516=2516 \times \frac{25}{16} = 25 and (54)2=5242=2516\left(\frac54\right)^2 = \frac{5^2}{4^2} = \frac{25}{16}⁠.) The 116\frac{1}{16} then comes out in front of the integral.

Using ∫dta2−t2=12alog⁡∣a+ta−t∣+C\int\frac{dt}{a^2 - t^2} = \frac{1}{2a}\log\left|\frac{a + t}{a - t}\right| + C with a=54a = \frac54⁠,

I=140[log⁡∣5+4t5−4t∣]−10=140[log⁡1−log⁡19]\begin{aligned} I &= \frac{1}{40}\left[\log\left|\frac{5 + 4t}{5 - 4t}\right|\right]_{-1}^0 \\ &= \frac{1}{40}\left[\log 1 - \log\frac19\right] \end{aligned}
Why?

The constant in front is

116⋅12a=116⋅25=140\frac{1}{16}\cdot\frac{1}{2a} = \frac{1}{16}\cdot\frac25 = \frac{1}{40}

Step by step, with a=54a = \frac54⁠:

2a=2×54=104=5212a=1÷52=25116⋅25=280=2÷280÷2=140\begin{aligned} 2a &= 2 \times \frac54 = \frac{10}{4} = \frac52 \\ \frac{1}{2a} &= 1 \div \frac52 = \frac25 \\[4pt] \frac{1}{16}\cdot\frac25 &= \frac{2}{80} \\ &= \frac{2 \div 2}{80 \div 2} = \frac{1}{40} \end{aligned}

Inside the log, multiply the top and the bottom by 4:

54+t54−t=5+4t5−4t\frac{\frac54 + t}{\frac54 - t} = \frac{5 + 4t}{5 - 4t}

At t=0t = 0 the fraction is 55=1\frac55 = 1⁠; at t=−1t = -1 it is 5−45+4=19\frac{5 - 4}{5 + 4} = \frac19⁠.

t=0:5+4(0)5−4(0)=55=1t=−1:5+4(−1)5−4(−1)  =5−45+4=19\begin{aligned} t = 0&: \frac{5 + 4(0)}{5 - 4(0)} = \frac55 = 1 \\[4pt] t = -1&: \frac{5 + 4(-1)}{5 - 4(-1)} \\ &\ \ = \frac{5 - 4}{5 + 4} = \frac19 \end{aligned}

Top limit minus bottom limit gives log⁡1−log⁡19\log 1 - \log\frac19⁠.

∴ I=140log⁡9\therefore\ I = \frac{1}{40}\log 9 or 120log⁡3\frac{1}{20}\log 3

Why?

log⁡1=0\log 1 = 0 and −log⁡19=log⁡9-\log\frac19 = \log 9⁠. Also log⁡9=log⁡32=2log⁡3\log 9 = \log 3^2 = 2\log 3⁠, which gives the second form.

log⁡19=log⁡1−log⁡9=0−log⁡9=−log⁡9log⁡1−log⁡19=0−(−log⁡9)=log⁡9140log⁡9=140(2log⁡3)=240log⁡3=120log⁡3\begin{aligned} \log\frac19 &= \log 1 - \log 9 \\ &= 0 - \log 9 = -\log 9 \\[4pt] \log 1 - \log\frac19 &= 0 - (-\log 9) \\ &= \log 9 \\[4pt] \frac{1}{40}\log 9 &= \frac{1}{40}(2\log 3) \\ &= \frac{2}{40}\log 3 = \frac{1}{20}\log 3 \end{aligned}

(b)

Let I=∫0π/2sin⁡2xtan⁡−1(sin⁡x) dxI = \displaystyle\int_0^{\pi/2}\sin 2x\tan^{-1}(\sin x)\,dx

=∫0π/22sin⁡xcos⁡xtan⁡−1(sin⁡x) dx= \displaystyle\int_0^{\pi/2}2\sin x\cos x\tan^{-1}(\sin x)\,dx⁠.

Why?

Writing sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x shows a function of sin⁡x\sin x multiplied by cos⁡x\cos x⁠, the derivative of sin⁡x\sin x⁠. That points to the substitution t=sin⁡xt = \sin x⁠.

Put sin⁡x=t\sin x = t⁠, so that cos⁡x dx=dt\cos x\,dx = dt⁠.

When x=0x = 0⁠, t=0t = 0⁠; when x=π2x = \frac{\pi}{2}⁠, t=1t = 1⁠.

Why?

sin⁡0=0\sin 0 = 0 and sin⁡π2=1\sin\frac{\pi}{2} = 1⁠. With the new limits we stay in tt to the end.

I=2∫01ttan⁡−1t dtI = 2\int_0^1 t\tan^{-1}t\,dt
Why?

Each piece is replaced: sin⁡x→t\sin x \to t⁠, tan⁡−1(sin⁡x)→tan⁡−1t\tan^{-1}(\sin x) \to \tan^{-1}t⁠, and cos⁡x dx→dt\cos x\,dx \to dt⁠. The 2 stays in front.

2 sin⁡x⏟t tan⁡−1(sin⁡x)⏟tan⁡−1t cos⁡x dx⏟dt=2 ttan⁡−1t dt\begin{aligned} &2\,\underbrace{\sin x}_{t}\,\underbrace{\tan^{-1}(\sin x)}_{\tan^{-1}t}\,\underbrace{\cos x\,dx}_{dt} \\ &= 2\,t\tan^{-1}t\,dt \end{aligned}

Integrating by parts, with tan⁡−1t\tan^{-1}t as the first function and tt as the second function,

Why?

The integrand is a product, so we integrate by parts:

∫u v dt=u∫v dt−∫(dudt∫v dt)dt\begin{aligned} \int u\,v\,dt &= u\int v\,dt \\ &\quad - \int\left(\frac{du}{dt}\int v\,dt\right)dt \end{aligned}

By ILATE, the inverse trig function tan⁡−1t\tan^{-1}t is the first function uu⁠. We can differentiate it easily, but we cannot easily integrate it.

I=2[tan⁡−1t⋅t22]01−2∫0111+t2⋅t22 dt\begin{aligned} I &= 2\left[\tan^{-1}t\cdot\frac{t^2}{2}\right]_0^1 \\ &\quad - 2\int_0^1\frac{1}{1 + t^2}\cdot\frac{t^2}{2}\,dt \end{aligned}
Why?

Here the two pieces are:

u=tan⁡−1t⇒dudt=11+t2v=t⇒∫v dt=t22\begin{aligned} u = \tan^{-1}t &\Rightarrow \frac{du}{dt} = \frac{1}{1 + t^2} \\ v = t &\Rightarrow \int v\,dt = \frac{t^2}{2} \end{aligned}
I=π4−∫011+t2−11+t2 dt=π4−∫01(1−11+t2)dt\begin{aligned} I &= \frac{\pi}{4} - \int_0^1\frac{1 + t^2 - 1}{1 + t^2}\,dt \\ &= \frac{\pi}{4} - \int_0^1\left(1 - \frac{1}{1 + t^2}\right)dt \end{aligned}
Why?

At t=1t = 1 the bracket is 2⋅π4⋅12=π42\cdot\frac{\pi}{4}\cdot\frac12 = \frac{\pi}{4}⁠; at t=0t = 0 it is 0. In the integral the 2 and the 12\frac12 cancel.

t=1:2⋅tan⁡−1(1)⋅(1)22  =2⋅π4⋅12=π4t=0:2⋅tan⁡−1(0)⋅(0)22  =2⋅0⋅0=0π4−0=π4\begin{aligned} t = 1&: 2\cdot\tan^{-1}(1)\cdot\frac{(1)^2}{2} \\ &\ \ = 2\cdot\frac{\pi}{4}\cdot\frac12 = \frac{\pi}{4} \\[4pt] t = 0&: 2\cdot\tan^{-1}(0)\cdot\frac{(0)^2}{2} \\ &\ \ = 2 \cdot 0 \cdot 0 = 0 \\[4pt] &\frac{\pi}{4} - 0 = \frac{\pi}{4} \end{aligned}

(tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4} because tan⁡π4=1\tan\frac{\pi}{4} = 1⁠.) In the integral:

2⋅11+t2⋅t22=t21+t22 \cdot \frac{1}{1 + t^2}\cdot\frac{t^2}{2} = \frac{t^2}{1 + t^2}

t21+t2\frac{t^2}{1 + t^2} has no standard integral as it stands, so add and subtract 1 in the numerator:

t21+t2=1−11+t2\frac{t^2}{1 + t^2} = 1 - \frac{1}{1 + t^2} t21+t2=1+t2−11+t2=1+t21+t2−11+t2=1−11+t2\begin{aligned} \frac{t^2}{1 + t^2} &= \frac{1 + t^2 - 1}{1 + t^2} \\ &= \frac{1 + t^2}{1 + t^2} - \frac{1}{1 + t^2} \\ &= 1 - \frac{1}{1 + t^2} \end{aligned}
I=π4−[t−tan⁡−1t]01=π4−(1−π4)\begin{aligned} I &= \frac{\pi}{4} - \Big[t - \tan^{-1}t\Big]_0^1 \\ &= \frac{\pi}{4} - \left(1 - \frac{\pi}{4}\right) \end{aligned}
Why?

Standard integrals: ∫1 dt=t\int 1\,dt = t and ∫dt1+t2=tan⁡−1t\int\frac{dt}{1 + t^2} = \tan^{-1}t⁠.

Upper limit minus lower limit: (1−π4)−(0−0)\left(1 - \frac{\pi}{4}\right) - (0 - 0)⁠. Watch the sign: −(−π4)=+π4-\left(-\frac{\pi}{4}\right) = +\frac{\pi}{4}⁠, so π4+π4=π2\frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}⁠.

[t−tan⁡−1t]01=(1−tan⁡−11)−(0−tan⁡−10)=(1−π4)−(0−0)=1−π4π4−(1−π4)=π4−1+π4=2π4−1=π2−1\begin{aligned} \Big[t - \tan^{-1}t\Big]_0^1 &= (1 - \tan^{-1}1) \\ &\quad - (0 - \tan^{-1}0) \\ &= \Big(1 - \frac{\pi}{4}\Big) - (0 - 0) \\ &= 1 - \frac{\pi}{4} \\[4pt] \frac{\pi}{4} - \Big(1 - \frac{\pi}{4}\Big) &= \frac{\pi}{4} - 1 + \frac{\pi}{4} \\ &= \frac{2\pi}{4} - 1 = \frac{\pi}{2} - 1 \end{aligned}

∴ I=π2−1\therefore\ I = \dfrac{\pi}{2} - 1⁠.

Where this comes from

Chapter 7, Integrals: Definite integrals by substitution and by parts.

Question type: (a) Substitution + Special forms (combined); (b) Substitution + Integration by parts (combined)

How to spot it: click to see

(a) 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 and 1x\frac{1}{x}⁠, or x2+1x^2 + 1 and xx⁠.

Here: First, the top sin⁡x+cos⁡x\sin x + \cos x is the derivative of sin⁡x−cos⁡x\sin x - \cos x⁠: put sin⁡x−cos⁡x=t\sin x - \cos x = t⁠. Squaring gives sin⁡2x=1−t2\sin 2x = 1 - t^2⁠, and the limits become −1-1 and 00⁠.

(a) Type 4. Special forms: complete the square, then match a card

Spot it: dxdx over a quadratic, or over the square root of a quadratic, with just a number on top.

Here: The integral left, ∫dt25−16t2\int\frac{dt}{25 - 16t^2}⁠, has only a number on top. Take 16 out to get (54)2−t2\left(\frac54\right)^2 - t^2 and use the a2−x2a^2 - x^2 card with a=54a = \frac54⁠.

(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 and 1x\frac{1}{x}⁠, or x2+1x^2 + 1 and xx⁠.

Here: First, writing sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x puts cos⁡x\cos x⁠, the derivative of sin⁡x\sin x⁠, beside tan⁡−1(sin⁡x)\tan^{-1}(\sin x)⁠: put sin⁡x=t\sin x = t⁠, with limits 00 and 11⁠.

(b) Type 8. Integration by parts: a product of two different kinds

Spot it: a product such as xsin⁡xx\sin x⁠, xexxe^x or xlog⁡xx\log x⁠. A lone log⁡x\log x or tan⁡−1x\tan^{-1}x works too: take the other function as 1.

Here: What is left, 2∫01ttan⁡−1t dt2\int_0^1 t\tan^{-1}t\,dt⁠, is a product: by ILATE take u=tan⁡−1tu = \tan^{-1}t and v=tv = t⁠.

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