Find the integrals of the functions- -tan 3 2 x 2 x d x

∫ tan^3 2x sec 2x dx. Let sec 2x =t ⇒ 2 sec 2x tan 2x =dt/dx⇒ dx =dt/2 sec 2x.tan 2x ∴∫ tan^3 2x sec 2x dx =∫ tan^3 2x sec 2x dt/2 sec 2x.tan 2x =1/2∫ tan^ ...

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Standard XII

Mathematics

Integration by Partial Fractions

Find the inte...

Question

Find the integrals of the functions.
$\int t a n^{3} 2 x s e c 2 x d x .$

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Solution

$\int t a n^{3} 2 x s e c 2 x d x . L e t s e c 2 x = t \Rightarrow 2 s e c 2 x t a n 2 x = \frac{d t}{d x} \Rightarrow d x = \frac{d t}{2 s e c 2 x . t a n 2 x} ∴ \int t a n^{3} 2 x s e c 2 x d x = \int t a n^{3} 2 x s e c 2 x \frac{d t}{2 s e c 2 x . t a n 2 x} = \frac{1}{2} \int t a n^{2} 2 x d t = \frac{1}{2} \int [s e c^{2} 2 x - 1] d t (∴ t a n^{2} x = s e c^{2} x - 1) = \frac{1}{2} \int [(t^{2} - 1) d t] = \frac{1}{2} [\frac{t^{3}}{3} - t] + C = \frac{1}{2} [\frac{s e c^{2} 2 x}{3} - s e c 2 x] + C = \frac{1}{6} s e c^{3} 2 x - \frac{1}{2} s e c 2 x + C$

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Find the integrals of the functions. ∫tan32x sec2xdx.

∫tan32x sec2xdx.Let sec2x=t⇒2sec2x tan2x=dtdx⇒dx=dt2sec2x.tan2x∴∫tan32x sec2xdx=∫tan32x sec2xdt2sec2x.tan2x=12∫tan22xdt=12∫[sec22x−1]dt (∴tan2x=sec2x−1)=12∫[(t2−1)dt]=12[t33−t]+C=12[sec22x3−sec2x]+C=16sec32x−12sec2x+C

Find the integrals of the functions.
$\int t a n^{3} 2 x s e c 2 x d x .$