Q.Integrate the following with respect to $x$:

Step 1. Part (i). $\dfrac{1}{\sin^2x}=\text{cosec}^2x$, and $\int\text{cosec}^2x\,dx=-\cot x+c$.

Step 2. Part (ii). $\dfrac{\tan x}{\cos x}=\dfrac{\sin x/\cos x}{\cos x}=\dfrac{\sin x}{\cos^2x}=\dfrac{\sin x}{\cos x}\cdot\dfrac{1}{\cos x}=\tan x\sec x=\sec x\tan x$, and $\int \sec x\tan x\,dx=\sec x+c$.

Step 3. Part (iii). $\dfrac{\cos x}{\sin^2x}=\dfrac{\cos x}{\sin x}\cdot\dfrac{1}{\sin x}=\cot x\cdot\text{cosec}\,x=\text{cosec}\,x\cot x$, and $\int \text{cosec}\,x\cot x\,dx=-\text{cosec}\,x+c$.

Step 4. Part (iv). $\dfrac{1}{\cos^2x}=\sec^2x$, and $\int\sec^2x\,dx=\tan x+c$.

Step 5. Check by differentiating. $\dfrac{d}{dx}(-\cot x)=\text{cosec}^2x$ ✓; $\dfrac{d}{dx}(\sec x)=\sec x\tan x$ ✓; $\dfrac{d}{dx}(-\text{cosec}\,x)=\text{cosec}\,x\cot x$ ✓; $\dfrac{d}{dx}(\tan x)=\sec^2x$ ✓.

Trig-identity simplification down to a standard derivative form

• Not recognising $\tan x/\cos x$ as $\sin x/\cos^2x=\sec x\tan x$ and instead attempting a harder route
• Sign slip between $\cot x$ and $-\cot x$, or between $\text{cosec}\,x$ and $-\text{cosec}\,x$