∫(cos 4x-1/cot x - tan x)dx is equal to

Question

Tardigrade Admin · Accepted Answer

I = ∫(cos 4x - 1/cot x - tan x) dx = ∫(-2 sin2 2x (sin x cos x)/(cos2 x - sin2 x)) dx =-∫ (sin2 2x sin 2x/cos 2x) x = ∫((cos2 2x -1)sin 2x/cos 2x) dx Let t = cos 2x or dt = -2 sin 2x dx . ∴ I = (1/2) ∫ ((1 - t2)/t) dt = (1/2) ln |t| - (t2/4) + C = (1/2) ln |cos 2x| - (1/4) cos2 2x + c

Q. $\int \frac{cos 4 x - 1}{co t x - t an x} d x$ is equal to

$\frac{1}{2} l n ∣ sec 2 x ∣ - \frac{1}{4} co s^{2} 2 x + c$

$\frac{1}{2} l n ∣ sec 2 x ∣ - + \frac{1}{4} co s^{2} 2 x + c$

$\frac{1}{2} l n ∣ cos 2 x ∣ - \frac{1}{4} co s^{2} 2 x + c$

$\frac{1}{2} l n ∣ cos 2 x ∣ + \frac{1}{4} co s^{2} 2 x + c$

Q. ∫cotx−tanxcos4x−1​dx is equal to

21​ln∣sec2x∣−41​cos22x+c

21​ln∣sec2x∣−+41​cos22x+c

21​ln∣cos2x∣−41​cos22x+c

21​ln∣cos2x∣+41​cos22x+c

I=∫cotx−tanxcos4x−1​dx =∫(cos2x−sin2x)−2sin22x(sinxcosx)​dx =−∫cos2xsin22xsin2x​x =∫cos2x(cos22x−1)sin2x​dx Let t=cos2x or dt=−2sin2xdx. ∴I=21​∫t(1−t2)​dt=21​ln∣t∣−4t2​+C =21​ln∣cos2x∣−41​cos22x+c

Q. $\int \frac{cos 4 x - 1}{co t x - t an x} d x$ is equal to

$\frac{1}{2} l n ∣ sec 2 x ∣ - \frac{1}{4} co s^{2} 2 x + c$

$\frac{1}{2} l n ∣ sec 2 x ∣ - + \frac{1}{4} co s^{2} 2 x + c$

$\frac{1}{2} l n ∣ cos 2 x ∣ - \frac{1}{4} co s^{2} 2 x + c$

$\frac{1}{2} l n ∣ cos 2 x ∣ + \frac{1}{4} co s^{2} 2 x + c$