∫ ( sin 2 x ⋅ sec 2 x+2 tan x ⋅ sin -1 x ⋅ √1-x2/√1-x2(1+ tan 2 x)) d x=

Question

∫ ( sin 2 x ⋅ sec 2 x+2 tan x ⋅ sin -1 x ⋅ √1-x2/√1-x2(1+ tan 2 x)) d x= (A) ( cos 2 x)( sin -1 x)+C (B) ( sin 2 x)( sin -1 x)+C (C) ( sec 2 x)( cos -1 x)+C (D) ( sec 2 x)( tan -1 x)+C

Tardigrade Admin · Accepted Answer

∫ ( sin 2 x ⋅ sec 2 x+2 tan x ⋅ sin -1 x ⋅ √1-x2/√1-x2(1+ tan 2 x)) d x =∫(( sin 2 x/√1-x2)+(2 tan x ⋅ sin -1 x/ sec 2 x)) d x =∫ ( sin 2 x/√1-x2) d x+∫ 2 sin x cos x sin -1 x d x = sin 2 x sin -1 x-∫ 2 sin x cos x sin -1 x d x+∫ 2 sin x cos x sin -1 x d x+ C = sin 2 x sin -1 x+C

Q. $\int \frac{s i n ^{2} x \cdot s e c ^{2} x + 2 t a n x \cdot s i n ^{- 1} x \cdot 1 - x ^{2}}{1 - x ^{2} ( 1 + t a n ^{2} x )} d x =$

$(cos^{2} x) (sin^{- 1} x) + C$

$(sin^{2} x) (sin^{- 1} x) + C$

$(sec^{2} x) (cos^{- 1} x) + C$

$(sec^{2} x) (tan^{- 1} x) + C$

Q. ∫1−x2​(1+tan2x)sin2x⋅sec2x+2tanx⋅sin−1x⋅1−x2​​dx=

(cos2x)(sin−1x)+C

(sin2x)(sin−1x)+C

(sec2x)(cos−1x)+C

(sec2x)(tan−1x)+C

∫1−x2​(1+tan2x)sin2x⋅sec2x+2tanx⋅sin−1x⋅1−x2​​dx =∫(1−x2​sin2x​+sec2x2tanx⋅sin−1x​)dx =∫1−x2​sin2x​dx+∫2sinxcosxsin−1xdx =sin2xsin−1x−∫2sinxcosxsin−1xdx+∫2sinxcosxsin−1xdx+C =sin2xsin−1x+C

Q. $\int \frac{s i n ^{2} x \cdot s e c ^{2} x + 2 t a n x \cdot s i n ^{- 1} x \cdot 1 - x ^{2}}{1 - x ^{2} ( 1 + t a n ^{2} x )} d x =$

$(cos^{2} x) (sin^{- 1} x) + C$

$(sin^{2} x) (sin^{- 1} x) + C$

$(sec^{2} x) (cos^{- 1} x) + C$

$(sec^{2} x) (tan^{- 1} x) + C$