Sample 2020-21 Class 12 Maths Q7 Solution: ∫ eˣ(1 − cot x + cosec²x) dx. Or: ∫ x² sin x dx from −π/2 to π/2

The question

(a) Find $\displaystyle\int e^x(1 - \cot x + \text{cosec}^2x)\,dx$

OR

(b) Evaluate $\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2\sin x\,dx$

Exam answer

(a)

Let $I = \int e^x(1 - \cot x + \text{cosec}^2x)\,dx$ ⁠.

Here $f(x) = 1 - \cot x$ and $f'(x) = \text{cosec}^2x$ ⁠.

Why?

An integrand of the form $e^x$ times (a function plus its derivative) has a ready-made answer, so we look for that pattern.

Since $\frac{d}{dx}(\cot x) = -\text{cosec}^2x$ ⁠, the derivative of $1 - \cot x$ is $0 - (-\text{cosec}^2x) = \text{cosec}^2x$ ⁠. So the bracket is $f(x) + f'(x)$ ⁠.

\begin{aligned} I &= \int e^x\{f(x) + f'(x)\}\,dx \\ &= e^x f(x) + C \end{aligned}

Why?

This standard result is the product rule read backwards:

\begin{aligned} &\frac{d}{dx}\left[e^x f(x)\right] \\ &= e^x f(x) + e^x f'(x) \end{aligned}

$\therefore\ I = e^x(1 - \cot x) + C$

(b)

Let $f(x) = x^2\sin x$ ⁠.

\begin{aligned} f(-x) &= (-x)^2\sin(-x) \\ &= -x^2\sin x = -f(x) \end{aligned}

Why?

The limits $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ are symmetric about 0, so we test whether $f$ is even or odd.

$(-x)^2 = x^2$ ⁠, and sine is odd: $\sin(-x) = -\sin x$ ⁠. So $f(-x) = -f(x)$ ⁠, and $f$ is odd.

\begin{aligned} (-x)^2 &= (-x)(-x) = x^2 \\ f(-x) &= (x^2)(-\sin x) \\ &= -x^2\sin x \end{aligned}

$\because f(x)$ is an odd function, $\int_{-a}^{a} f(x)\,dx = 0$ ⁠.

Why?

For an odd function, the area below the axis on one side of 0 cancels the area above it on the other side. Here $a = \frac{\pi}{2}$ ⁠.

Watch out: the property needs limits of the form $-a$ to $a$ ⁠. For an even function it gives $2\int_0^a f(x)\,dx$ instead.

$\therefore \displaystyle\int_{-\pi/2}^{\pi/2} x^2\sin x\,dx = 0$

Where this comes from

Chapter 7, Integrals: ∫eˣ(f(x) + f′(x))dx; integral of an odd function.

Question type: (a) The $e^x$ shortcut; (b) Limits from $-a$ to $a$

How to spot it: click to see

(a) Type 9. The $e^x$ shortcut: a function and its derivative in a bracket

Spot it: $e^x$ times a bracket of two parts, where one part is the derivative of the other.

Here: The bracket holds $1 - \cot x$ and $\text{cosec}^2x$ ⁠, and $\text{cosec}^2x$ is the derivative of $1 - \cot x$ ⁠. So take $f(x) = 1 - \cot x$ and write $e^x(1 - \cot x) + C$ at once.

(b) Type 11. Limits from $-a$ to $a$ ⁠: check odd or even

Spot it: the two limits are the same number with opposite signs.

Here: The limits are $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ ⁠. Putting $-x$ leaves $x^2$ alone but flips the sign of $\sin x$ ⁠, so $x^2\sin x$ is odd and the integral is 0.

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Q7: ∫ex(1−cot⁡x+cosec2x) dx\int e^x(1 - \cot x + \text{cosec}^2x)\,dx∫ex(1−cotx+cosec2x)dx⁠. Or: ∫−π/2π/2x2sin⁡x dx\int_{-\pi/2}^{\pi/2} x^2\sin x\,dx∫−π/2π/2​x2sinxdx

Q7: $\int e^x(1 - \cot x + \text{cosec}^2x)\,dx$ ⁠. Or: $\int_{-\pi/2}^{\pi/2} x^2\sin x\,dx$