`intcos2thetalog(1+tantheta)dx`

Solve: (1-tantheta)(1+sin2theta)=(1+tantheta)

The value of (2cos^(2)theta-1)((1+tantheta)/(1-tantheta)+(1-tantheta)/(1+tantheta)) is

int_(sintheta)^(costheta)f(xtantheta)dx(w h e r etheta!=(npi)/2,n in I))i se q u a lto (a) -costhetaint_1^(tantheta)f(xsintheta)dx (b)-tanthetaint_(costheta)^(sintheta)f(x)dx (c)sinthetathetaint_1^(tantheta)f(xcostheta)dx (d)1/(tantheta)thetaint_(sintheta)^(sinthetatantheta)f(x)dx

(2sinthetatantheta(1-tantheta)+2sinthetasec^(2)theta)/((1+tantheta)^(2))=?

(tantheta)/(1-cottheta)+(cottheta)/(1-tantheta) is equal to -

(tantheta)/(1-cottheta)+(cottheta)/(1-tantheta)=1+tantheta+cottheta=sectheta"cosec"theta+1

(tantheta+2)(2 tantheta+1)= ?

intcos2xlogsinxdx

costheta(tantheta+2)(2tantheta+1)=?