19.sin x cos3x

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Solution

The given function is 1 sinx cos 3 x .

The given function can be written as,

1 sinx cos 3 x = sin 2 x+ cos 2 x sinx cos 3 x = sinx cos 3 x + 1 sinxcosx =tanx sec 2 x+ 1 cos 2 x sinxcosx cos 2 x =tanx sec 2 x+ sec 2 x tanx (1)

From (1), we get,

∫ 1 sinx cos 3 x dx = ∫ ( tanx sec 2 x )dx + ∫ [ 1 cos 2 x sinxcosx cos 2 x ] dx = ∫ ( tanx sec 2 x )dx + ∫ sec 2 x tanx dx (2)

Put tanx=t sec 2 xdx=dt

Substitute t for tanx and dt for sec 2 xdx in (1),

∫ 1 sinx cos 3 x dx = ∫ tdt+ ∫ 1 t dt = t 2 2 +log( t ) = tan 2 x 2 +log(tanx)+c

Thus, the integral of the function 1 sinx cos 3 x is tan 2 x 2 +log(tanx)+c.

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Similar questions

\int \frac{d x}{sin x {cos}^{3} x}

\int \frac{sin x {cos}^{3} x}{1 + {cos}^{2} x} d x

\frac{1}{sin x {cos}^{3} x}

f_{n} (x) = \frac{sin x}{cos 3 x} + \frac{sin 3 x}{cos 3^{2} x} + \frac{sin 3^{2} x}{cos 3^{3} x} + \dots + \frac{sin 3^{n - 1} x}{cos 3^{n} x}

f_{2} (\frac{π}{4}) + f_{3} (\frac{π}{4}) =

s i n^{3} x c o s x + s i n^{2} x c o s^{2} x + s i n x c o s^{3} x = 1

[0, 2 π]

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