Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 72 Marks
Question
Evaluate: $\int \sec ^{\frac{4}{3}} x \csc ^{\frac{8}{3}} x d x$
✓
Answer
Let $I =\int \sec ^{\frac{4}{3}} x \csc ^{\frac{8}{3}} x \ d x$. Then, we have
$I=\int \frac{1}{\cos ^{4 / 3} x \sin ^{8 / 3} x} d x=\int \cos ^{\frac{-4}{3}} x \sin ^{\frac{-8}{3}} x\ d x$
since $-\left(\frac{4}{3}+\frac{8}{3}\right)=-4$, which is an even integer.
So, we divide both numerator and denominator by $\cos ^4 x$.
$\therefore I=\int \frac{\sec ^4 x}{\tan ^{8 / 3} x} d x=\int \frac{\left(1+\tan ^2 x\right)}{\tan ^{8 / 3} x} \sec ^2 x dx$
Put $\tan x=t$ and $\sec ^2=d t$, we get
$I=\int \frac{1+t^2}{t^{\frac{8}{3}}} d t=\int\left(t^{\frac{-8}{3}}+t^{\frac{-2}{3}}\right) d t=-\frac{3}{5} t^{\frac{-5}{3}}+3 t^{\frac{1}{3}}+c$
$\Rightarrow I=-\frac{3}{5} \tan ^{\frac{-5}{3}} x+3 \tan ^{\frac{1}{3}} x+C$
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