Sin 3 x-cos 4 x-3 cos 2 x-1 tan -1 x-cos x d x-

The correct option is A log |tan^ 1 (sec x + cos x)| + cPut tan^ 1 (sec x + cos x) = f(x) f^1(x) = sin^3 x/cos^4 x + 3 cos^2 x+1 ∴∫f^1 (x)/f(x) = log |f(x)|+c ...

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Byju's Answer

Standard XII

Mathematics

Standard Formulae - 2

sin 3 x/cos 4...

Question

$\frac{s i n^{3} x}{(c o s^{4} x + 3 c o s^{2} x + 1) t a n^{- 1} (s e c x + c o s x)} d x =$

t a n^{- 1} (s e c x + c o s x) + c

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l o g | t a n^{- 1} (s e c x + c o s x) | + c

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\frac{1}{(s e c x + c o s^{2} x)^{2}} + c

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l o g | s e c x + c o s x | + c

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Solution

The correct option is B $l o g | t a n^{- 1} (s e c x + c o s x) | + c$
Put $t a n^{- 1} (s e c x + c o s x) = f (x) f^{1} (x) = \frac{s i n^{3} x}{c o s^{4} x + 3 c o s^{2} x + 1} ∴ \int \frac{f^{1} (x)}{f (x)} = l o g | f (x) | + c$

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Similar questions

\int \frac{{sin}^{3} x}{({cos}^{4} x + 3 {cos}^{2} x + 1) {tan}^{- 1} (sec x + cos x)} d x =

Q. Evaluate:

\int \frac{{sin}^{3} x}{({cos}^{4} x + 3 {cos}^{2} x + 1) {tan}^{- 1} (sec x + cos x)} d x

\int \frac{{sin}^{3} x}{({cos}^{4} x + 3 {cos}^{2} x + 1) {tan}^{- 1} (sec x + cos x)} d x

is equal to.

\frac{{sin}^{3} x}{({cos}^{4} x + 3 {cos}^{2} x + 1) {tan}^{- 1} (sec x + cos x)} d x =

\int \frac{{sin}^{3} x}{({cos}^{4} x + 3 {cos}^{2} x + 1) {tan}^{- 1} (sec x + cos x)} d x =

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sin3x(cos4x+3cos2x+1)tan−1(secx+cosx)dx=

The correct option is B log|tan−1(secx+cosx)|+cPut tan−1(secx+cosx)=f(x)f1(x)=sin3xcos4x+3cos2x+1∴∫f1(x)f(x)=log|f(x)|+c

$\frac{s i n^{3} x}{(c o s^{4} x + 3 c o s^{2} x + 1) t a n^{- 1} (s e c x + c o s x)} d x =$

The correct option is B $l o g | t a n^{- 1} (s e c x + c o s x) | + c$
Put $t a n^{- 1} (s e c x + c o s x) = f (x) f^{1} (x) = \frac{s i n^{3} x}{c o s^{4} x + 3 c o s^{2} x + 1} ∴ \int \frac{f^{1} (x)}{f (x)} = l o g | f (x) | + c$