Plz tell which one to take as substituent in each of the questions also plz explain how we decide which - Maths - Integrals

Question

Plz tell which one to take as substituent in each of the
questions also plz explain how we decide which one to be
substituted as t in such questions
19 Integrate : ∫ sinx - cosx 1   -   sin2x
dx
20 Integrate: ∫ sin2x 3   sin 2 x + 4 cos 2
x   dx

Lovina Kansal · Accepted Answer

Dear student
we have, I=∫sin2x3sin2x+4cos2xdx=∫sin2x3sin2x+41-sin2xdx=∫sin2x3sin2x+4-4sin2xdx=∫sin2x4-sin2xdx=-∫sin2xsin2x-4dx=-∫2sinx cosx1-cos2x-4dx=∫2cosxcos2x+3sinx dxtake u=cosx⇒du=-sinx dxI=-2∫uu2+3duTake u2+3=v⇒2udu=dvSo, I=-∫1vdv=-lnv+C=-lnu2+3+C=-lncos2x+3+C
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Plz tell which one to take as substituent in each of the questions also plz explain how we decide which one to be substituted as t in such questions 19. Integrate : ∫ sinx - cosx 1 - sin2x dx 20. Integrate: ∫ sin2x 3 sin 2 x + 4 cos 2 x dx

Plz tell which one to take as substituent in each of the questions also plz explain how we decide which one to be substituted as t in such questions
19. Integrate : $\int \frac{sinx - cosx}{\sqrt{1 - sin2x}} dx$
20. Integrate: $\int \frac{sin2x}{3 \sin^{2} {x + 4 cos}^{2} x} dx$