Q.Find the following integrals:

Concept: Use standard trigonometric integrals and algebraic simplification before integrating.

(i) $\int (\sin x + \cos x)\, dx$

Integrate term‑by‑term:

$\int \sin x\, dx = -\cos x$, $\int \cos x\, dx = \sin x$.

So the result is $-\cos x + \sin x + C$.

(ii) $\int \operatorname{cosec} x\,(\operatorname{cosec} x + \cot x)\, dx$

Expand: $\int (\operatorname{cosec}^2 x + \operatorname{cosec} x \cot x)\, dx$.

Standard integrals: $\int \operatorname{cosec}^2 x\, dx = -\cot x$, $\int \operatorname{cosec} x \cot x\, dx = -\operatorname{cosec} x$.

Thus $-\cot x - \operatorname{cosec} x + C$.

(iii) $\int \dfrac{1 - \sin x}{\cos^2 x}\, dx$ …

The core idea here is that integration is the reverse of differentiation. When you see a sum inside an integral, you can break it apart — the integral of a sum is the sum of the integrals. Then you just need to recognise each piece as the derivative of some known function.

Why the Power Rule matters here: The Power Rule says $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$. But in trigonometry, we often use its “disguised” form: if you know that $\frac{d}{dx}(\sin x) = \cos x$, then $\int \cos x \, dx = \sin x + C$ is just the Power Rule applied to the function $\sin x$ as a “variable”. Same logic works for all six trigonometric functions.

Let’s work through each part.

(i) $\int (\sin x + \cos x)\, dx$

1. Split the sum:

   $\int (\sin x + \cos x)\, dx = \int \sin x \, dx + \int \cos x \, dx$

2. Recall the basic derivatives:

   $\frac{d}{dx}(\cos x) = -\sin x$, so $\int \sin x \, dx = -\cos x + C_1$

   $\frac{d}{dx}(\sin x) = \cos x$, so $\int \cos x \, dx = \sin x + C_2$

3. Combine constants:

   $-\cos x + \sin x + C$ (where $C = C_1 + C_2$)

(ii) $\int \operatorname{cosec} x\,(\operatorname{cosec} x + \cot x)\, dx$

1. Expand the product:

   $\operatorname{cosec} x \cdot \operatorname{cosec} x + \operatorname{cosec} x \cdot \cot x = \operatorname{cosec}^2 x + \operatorname{cosec} x \cot x$

2. Recognise standard derivatives:

   $\frac{d}{dx}(\cot x) = -\operatorname{cosec}^2 x$, so $\int \operatorname{cosec}^2 x \, dx = -\cot x + C_1$

   $\frac{d}{dx}(\operatorname{cosec} x) = -\operatorname{cosec} x \cot x$, so $\int \operatorname{cosec} x \cot x \, dx = -\operatorname{cosec} x + C_2$

3. Add them up:

   $\int (\operatorname{cosec}^2 x + \operatorname{cosec} x \cot x)\, dx = -\cot x - \operatorname{cosec} x + C$

(iii) $\int \dfrac{1 - \sin x}{\cos^2 x}\, dx$ …

Method: Simplify with Trig Identities, Then Integrate Standard Forms

Use this for integrals of trigonometric products or quotients that are not yet standard forms: rearrange the integrand algebraically or with an identity until each piece is a known antiderivative.

Steps

Step 1: Expand or simplify the integrand first.

Multiply out products such as $\operatorname{cosec}x(\operatorname{cosec}x+\cot x)=\operatorname{cosec}^2x+\operatorname{cosec}x\cot x$, or split a quotient like $\dfrac{1-\sin x}{\cos^2 x}=\sec^2 x-\sec x\tan x$.

Step 2: Match each term to a standard antiderivative.

Rely on the reverse of standard derivatives: …

Common Mistakes

Mistake 1: Sign error $\int\sin x\,dx = \cos x$.

Why it's wrong: $\frac{d}{dx}\cos x=-\sin x$, so $\int\sin x\,dx=-\cos x+C$. Correct approach: remember integrating sine introduces a minus sign.

Mistake 2: Trying to integrate $\dfrac{1-\sin x}{\cos^2 x}$ as a quotient directly.

Why it's wrong: there is no quotient rule for integrals. Correct approach: split it into $\sec^2x-\sec x\tan x$ first, which are standard forms. …