First slide

Introduction to integration

Question

If polynomials $P$ and $Q$ satisfies
$\begin{array}{l} \int ((3 x - 1) \cos x + (1 - 2 x) \sin x) d x \\ = P \cos x + Q \sin x \end{array}$
$(ignoring the constant of integration) then$

Easy

A

$P = 3 x - 12$
B

$Q = 2 + x$
C

$P = 3 (x - 1)$
D

$Q = 3 (x - 1)$

Solution

Integrating by parts, the given integral is equal to
$\begin{array}{l} (3 x - 1) \sin x - 3 \int \sin x d x - (1 - 2 x) \cos x - 2 \int \cos x d x \\ = (3 x - 1) \sin x + 3 \cos x - (1 - 2 x) \cos x - 2 \sin x \\ = (3 x - 3) \sin x + (2 + 2 x) \cos x \end{array}$
Hence $Q = 3 (x - 1)$ and $P = 2 (1 + x)$

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