∫ ( sin 2 x+ sin 4 x- sin 6 x/1+ cos 2 x+ cos 4 x+ cos 6 x) d x is equal to (where C is indefinite integration constant.)

Question

Tardigrade Admin · Accepted Answer

I=∫ ( sin 2 x+ sin 4 x- sin 6 x/1+ cos 2 x+ cos 4 x+ cos 6 x) d x=∫ (2 sin 3 x cos x-2 sin 3 x cos 3 x/2 cos 2 x+2 cos x cos 5 x) d x =∫ ( sin 3 x( cos x- cos 3 x)/ cos x( cos x+ cos 5 x)) d x=∫ (2 sin 3 x sin 2 x sin x/2 cos x cos 2 x cos 3 x) d x=∫ tan x tan 2 x tan 3 x d x Θ 3 x=2 x+x ⇒ tan 3 x=( tan 2 x+ tan x/1- tan 2 x tan x) ⇒ tan 3 x- tan 3 x tan 2 x tan x= tan 2 x+ tan x ⇒ tan 3 x tan 2 x tan x= tan 3 x- tan 2 x- tan x ∴ I=∫( tan 3 x- tan 2 x- tan x) d x=(1/3) ln | sec 3 x|-(1/2) ln | sec 2 x|- ln | sec x|+C

Q. $\int \frac{s i n 2 x + s i n 4 x - s i n 6 x}{1 + c o s 2 x + c o s 4 x + c o s 6 x} d x$ is equal to (where $C$ is indefinite integration constant.)

$\frac{1}{3} ln ∣ sec 3 x ∣ + \frac{1}{2} ln ∣ sec 2 x ∣ + ln ∣ sec x ∣ + C$

$\frac{1}{3} ln ∣ sec 3 x ∣ - \frac{1}{2} ln ∣ sec 2 x ∣ - ln ∣ sec x ∣ + C$

$C - \frac{1}{3} ln ∣ sec 3 x ∣ + \frac{1}{2} ln ∣ sec 2 x ∣ + ln ∣ sec x ∣$

$C - \frac{1}{3} ln ∣ sec 3 x ∣ - \frac{1}{2} ln ∣ sec 2 x ∣ - ln ∣ sec x ∣$

Q. ∫1+cos2x+cos4x+cos6xsin2x+sin4x−sin6x​dx is equal to (where C is indefinite integration constant.)

31​ln∣sec3x∣+21​ln∣sec2x∣+ln∣secx∣+C

31​ln∣sec3x∣−21​ln∣sec2x∣−ln∣secx∣+C

C−31​ln∣sec3x∣+21​ln∣sec2x∣+ln∣secx∣

C−31​ln∣sec3x∣−21​ln∣sec2x∣−ln∣secx∣

Q. $\int \frac{s i n 2 x + s i n 4 x - s i n 6 x}{1 + c o s 2 x + c o s 4 x + c o s 6 x} d x$ is equal to (where $C$ is indefinite integration constant.)

$\frac{1}{3} ln ∣ sec 3 x ∣ + \frac{1}{2} ln ∣ sec 2 x ∣ + ln ∣ sec x ∣ + C$

$\frac{1}{3} ln ∣ sec 3 x ∣ - \frac{1}{2} ln ∣ sec 2 x ∣ - ln ∣ sec x ∣ + C$

$C - \frac{1}{3} ln ∣ sec 3 x ∣ + \frac{1}{2} ln ∣ sec 2 x ∣ + ln ∣ sec x ∣$

$C - \frac{1}{3} ln ∣ sec 3 x ∣ - \frac{1}{2} ln ∣ sec 2 x ∣ - ln ∣ sec x ∣$