CBSE 2025 · Region 4 · Set 1 · Q34 · 5 marks
Find : \[\int \frac{\cos x}{\left(4+\sin ^{2} x\right)\left(5-4 \cos ^{2} x\right)} d x \]Evaluate : \[\int_{0}^{\pi} \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} \]
Find : \[\int \frac{\cos x}{\left(4+\sin ^{2} x\right)\left(5-4 \cos ^{2} x\right)} d x \]
Evaluate : \[\int_{0}^{\pi} \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} \]
Marking-scheme solution
\[\begin{aligned}
I & =\int \frac{\cos x}{\left(4+\sin ^{2} x\right)\left(5-4 \cos ^{2} x\right)} d x \\
& =\int \frac{\cos x}{\left(4+\sin ^{2} x\right)\left(1+4 \sin ^{2} x\right)} d x
\end{aligned}
\]
\[\begin{aligned}
I & =\int \frac{d t}{\left(4+t^{2}\right)\left(1+4 t^{2}\right)} \\
& =-\frac{1}{15} \int \frac{d t}{4+t^{2}}+\frac{4}{15} \int \frac{d t}{1+4 t^{2}} \quad(\because \text { using Partial Fraction }) \\
& =-\frac{1}{30} \tan ^{-1}\left(\frac{t}{2}\right)+\frac{2}{15} \tan ^{-1}(2 t)+C \\
& =-\frac{1}{30} \tan ^{-1}\left(\frac{\sin x}{2}\right)+\frac{2}{15} \tan ^{-1}(2 \sin x)+C
\end{aligned}
\]
\[\begin{aligned}
& I=\int_{0}^{\pi} \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} \\
&=2 \int_{0}^{\pi / 2} \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} \\
&=2 \int_{0}^{\pi / 2} \frac{\sec ^{2} x}{a^{2}+b^{2} \tan ^{2} x} d x \\
& \tan x=t \text { gives } \\
& I=2 \int_{0}^{\infty} \frac{d t}{a^{2}+b^{2} t^{2}} \\
&\left.=\frac{2}{b^{2}} \cdot \frac{b}{a} \tan ^{-1}\left(\frac{b t}{a}\right)\right]_{0}^{\infty} \\
&=\frac{\pi}{a b}
\end{aligned}
\]
IntegralsMethods of IntegrationApplylong_answerhard
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.