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Mathematics · 2024 · 5 marks
CBSE 2024 · Region 1 · Set 3 · Q34
(a) Find: ∫ ((3 cos x -2) sin x)/(5- sin^2 x -4 cos x) d x
(a)
Find : \[\int \frac{(3 \cos \mathrm{x}-2) \sin \mathrm{x}}{5-\sin ^{2} \mathrm{x}-4 \cos \mathrm{x}} d \mathrm{x} \]
OR
(b)
Evaluate : \[\int_{-2}^{2} \frac{\mathrm{x}^{3}+|\mathrm{x}|+1}{\mathrm{x}^{2}+4|\mathrm{x}|+4} d \mathrm{x} \]
Marking-scheme solution
$$\begin{aligned}
& \int \frac{(3 \cos \mathrm{x}-2) \sin \mathrm{x}}{5-\sin ^{2} \mathrm{x}-4 \cos \mathrm{x}} d \mathrm{x}, \quad \text { Put } \cos \mathrm{x}=\mathrm{t} \text { so that, }-\sin \mathrm{x} \mathrm{dx}=\mathrm{dt} \\
& =\int \frac{2-3 \mathrm{t}}{5-\left(1-\mathrm{t}^{2}\right)-4 \mathrm{t}} d \mathrm{t} \\
& =\int \frac{2-3 \mathrm{t}}{(\mathrm{t}-2)^{2}} d \mathrm{t} \\
& \int \frac{2-3 \mathrm{t}}{(\mathrm{t}-2)^{2}} d \mathrm{t}=-3 \int \frac{1}{\mathrm{t}-2} d \mathrm{t}-4 \int \frac{1}{(\mathrm{t}-2)^{2}} \mathrm{dt} \\
& \quad=-3 \log |\mathrm{t}-2|-4\left(\frac{-1}{\mathrm{t}-2}\right)+C \\
& \quad=-3 \log |\cos \mathrm{x}-2|+\frac{4}{\cos \mathrm{x}-2}+C
\end{aligned}
$$
$\displaystyle \mathrm{I}=\int_{-2}^{2} \frac{\mathrm{x}^{3}+|\mathrm{x}|+1}{\mathrm{x}^{2}+4|\mathrm{x}|+4} d \mathrm{x}$
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CBSE Class 12 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.