`int (cos 2x-1)/(cos 2x+1)dx`は次と等しい

#BSEB2021||#Indefnite_integration||Integration of 1-cos2x/1+cos2x dx||#shorts

`intsqrt(1-cos 2x) dx`

積分 ∫ cos2x/(1-sin2x) - √(​​1-x) dx - sin2x/cos2x dx - 24(1-x)^5 dx - cos⁡(2-x/3)- 5x/(x^2-25)

If ∫√sec 2x-1dx= loge|cos 2x + β +√cos 2x (1+cos 1/β x| + constant, then β- α is equal to

If Integral of √sec 2x-1 dx=a log e|cos 2x+b+√cos 2x(1+cos 1/b x)| + constant, then b-a is equal to:

(1 - cos 2x)/(1 + cos 2x) の積分

integrate (cos x - cos 2x)/(1 - cos x) dx □□□ int cos x-cos*2x 1-cos x dx

 If ∫sec2x-1dx=αlogecos2x+β+cos2x1+cos1βxconstant, then &beta....

integration of Cos^2(x) #integration #math #calculus #integrationproblems #mathmagic

Integrate the functions 25. 1/cos^2 x (1-tan^2x) || Ex 7.2 Class 12 Mathematics NCERT Solution

If integration of root over (sec2x-1) dx= a log|(cos2x+b)+root(cos2x(1+cosx/b))+constant, then b- a=

If integral of sqrt(2x-1)dx = aln(cos2x + b + sqrt(cos2x(1+cos()x/b))), then, b-a=

Integration of (cosx-cos2x)/(1-cosx)

Evaluate: ∫ (cos 2x/ sin square x .cos square x) dx

Integration of (cosx-cos2x)/(1-cosx) dx

∫(sin(2x) - sin²(x))/(cos(2x) - cos²(x)) dx. MIT Integration Bee 2018, Question 15, Qualifying Exam.

Integrate the function ∫ e^2x (1+sin2x/1+cos2x) dx

Integration of (cosx-cos2x)/(1-cosx) | ∫ (cosx-cos2x)/(1-cosx) dx

Find integral dx/cos^2x(1-tanx)^2 | Class12 | integrals | NCERT Maths