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JAMB Mathematics 2014 Paper

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Question : 32

Total: 47

Evaluate

\int \sin 2x \, dx

cos 2x + k
$\frac{1}{2}$ cos 2x + k
$-\frac{1}{2}$ cos 2x + k
-cos 2x + k

Solution:

Concept:

The derivative of

-\cos(2x)

gives a factor of

2

, so the integral needs a reciprocal factor of

\frac{1}{2}

.

Explanation:

Let

u = 2x

, so

du = 2\,dx

, giving

dx = \frac{du}{2}

.

Substitute into the integral:

\int \sin 2x \, dx = \int \sin u \cdot \frac{du}{2}

.

Integrate with respect to

u

:

\frac{1}{2}(-\cos u) + k

.

Replace

u

with

2x

to get

-\frac{1}{2}\cos 2x + k

.

This matches option C.

Answer:

C.

-\frac{1}{2}\cos 2x + k

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