V dx d dx d dx d dx d dx -sec-¹(x) tan-¹(x) cos-¹(x) csc-¹(x) d dx sin ¹(x) cot-¹(x) 1. 2. 3. 4. 5. 6. 8. assa 9. H M 7. H TARA - BEOGEN - - arcsin (r) darctan (x) arcsec (x) dx d dx 1+x² 1 |x|√x²-1 COS X 1 sec²x 1 sec(x) tan(x)

Q: 1. Dx[arcsin(2x + 1)]

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Q: y=(sinx)arctanx

A:

Q: Scos x vsin x dx (sin x)i-(sinx) +C (sin x)i +(sinx)+c c. (sinx)- (sinx) + C (sinx)i +(sinx)+c A. B.…

A: ∫cos3xsinxdx

Q: (tan x) = fu. v dx = (sin x) = | -=-=dx = (cos x) = fe" dx = | (arcsinx) = sec x tan x dx = secx tan…

A: We have to find out derivative and integral

Q: se*cosx dx e-sinx

A: Let sinx = u then cosxdx = du

Q: cos x- sin dx 1- sin 2x

A:

Q: 8.) k(x)=-2 tan(2(x+)) Period = Increment =

A: Given function is: kx=-2tan2x+π8 Then the period of this function is 2π2=π. The increment is π8.

Q: sin 2x (1 - 2 cos 2x)? dx 6.

A:

Q: u) sec(2-x)tan(2-x)dx

A: To evaluate ∫sec2-xtan2-xdx

Q: COs 2x cotx x – tanx %3D sinx · cOsx

A:

Q: If  arcsin(x) = 1.25  then  arcsin(-x) = ?

A: It is given that, arcsin x=1.25 Take sine of both sides: sin arcsin x=sin 1.25 Result: sin arcsin…

Q: sec x tanx (1+sec'x)² + sec

A: substitute u= sec3(x)+1 du/dx=3sec3(x)tan(x) dx = du/(3sec3(x)tan(x)) put this in equation  1/3∫…

Q: Find dy. y=tan-1(ex^10)   dy=(______) dx

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Q: cos x-sinxtanx=

A:

Q: cos(2x) dx sinx+cosx dx Vex

A: Find your answer below

Q: Stan (x)sec°(x)dx

A: Since you have asked multiple question, we will solve the first question for you as per our guide…

Q: sec² (x) tan° (x)dx

A: