Question

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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$\huge\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Answer}}}}}}}} \\ \large\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Your~answer↓}}}}}}}}$

$∫  \frac{1}{cos2x + {3sin}^{2} x} dx \\ =  ∫  \frac{1}{1 - 2 {sin}^{2} x + {3sin}^{2}x } dx \\ =  ∫  \frac{1}{1 + {sin}^{2} x} dx \\ \small\bold\red{(divide \: numerator \: and \: denominator \: by \: {cos}^{2} x)} \\ =  ∫  \frac{ {sec}^{2}x }{ \: {sec}^{2}x \: + {tan}^{2} x } dx \\ =  ∫  \frac{ {sec}^{2}x }{1 + {tan}^{2}x + {tan}^{2} x } dx \\ =  ∫  \frac{ {sec}^{2}x }{1 + {2tan}^{2} x} dx \\ \small\bold\red{(put \: tanx \: = t \: = > \: {sec}^{2} x \: dx \: = dt)} \\ =  ∫  \frac{1}{1 + {2t}^{2} } dt \\ = \frac{1}{2}  ∫  \frac{dt}{ {t}^{2} + \frac{1}{2} } \\ = \frac{1}{2}  ∫  \frac{dt}{ {t}^{2} + {( \frac{1}{ \sqrt{2} } )}^{2} } \\ = \frac{1}{2} \times \frac{1}{ \frac{1}{ \sqrt{2} } } {tan}^{ - 1} \frac{t}{ \frac{1}{ \sqrt{2} } } + c \\ = \frac{1}{ \sqrt{2} } {tan}^{ - 1} \sqrt{2} t + c \\ = \frac{1}{ \sqrt{2} } {tan}^{ - 1} ( \sqrt{2} tanx) + c$

$\large\bold\red{formula \: used} \\ \large\bold\red{cos2x = 1 - {2sin}^{2} x} \\ \large\bold\red{ ∫  \frac{dx}{ {x}^{2} + {a}^{2} } = \frac{1}{a} {tan}^{ - 1} \frac{x}{a} + c}$

$\huge \fcolorbox{black}{cyan}{♛hope \: it \: helps \: you♛}$