Q.
Match the following function in column I with their anti-derivatives in column II and choose the correct option from the codes given below.
Column I
Column II
A
$\cos x$
1
$-\cot x+C$
B
$\sin x$
2
$\sin x+C$
C
$\sec ^2 x$
3
$-\cos x+C$
D
$\operatorname{cosec}^2 x$
4
$\tan x+C$
| Column I | Column II | ||
|---|---|---|---|
| A | $\cos x$ | 1 | $-\cot x+C$ |
| B | $\sin x$ | 2 | $\sin x+C$ |
| C | $\sec ^2 x$ | 3 | $-\cos x+C$ |
| D | $\operatorname{cosec}^2 x$ | 4 | $\tan x+C$ |
Integrals
Solution:
Derivatives
Anti - Derivatives
A
$\frac{d}{d x}(\sin x)=\cos x $;
$ \int \cos x d x=\sin x+C$
B
$\frac{d}{d x}(-\cos x)=\sin x$;
$ \int \sin x d x=-\cos x+C$
C
$\frac{d}{d x}(\tan x)=\sec ^2 x$;
$ \int \sec ^2 d x=\tan x+C$
D
$\frac{d}{d x}(-\cot x)=\operatorname{cosec}^2 x$;
$ \int \operatorname{cosec}^2 x d x=-\cot x+C$
| Derivatives | Anti - Derivatives | |
|---|---|---|
| A | $\frac{d}{d x}(\sin x)=\cos x $; | $ \int \cos x d x=\sin x+C$ |
| B | $\frac{d}{d x}(-\cos x)=\sin x$; | $ \int \sin x d x=-\cos x+C$ |
| C | $\frac{d}{d x}(\tan x)=\sec ^2 x$; | $ \int \sec ^2 d x=\tan x+C$ |
| D | $\frac{d}{d x}(-\cot x)=\operatorname{cosec}^2 x$; | $ \int \operatorname{cosec}^2 x d x=-\cot x+C$ |