Question

Please solve this ASAP
and please no fake answers
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Answer 1

Answer:

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Answer 2

Question

$\sf\dfrac{sinx}{ {a}^{2} + {b}^{2} {cos}^{2} x}$

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

$= > \displaystyle \int \dfrac{sinx}{ {a}^{2} + {b}^{2} {cos}^{2}x } dx$

Let bcosx=dt

-bsinxdx=dt

$\sf \: sinxdx = - \dfrac{dt}{b}$

$\sf = > - \dfrac{1}{b} \displaystyle\int \dfrac{dt}{ {a}^{2} + {t}^{2} }$

$\sf \boxed{\displaystyle\int \dfrac{dx}{ {a}^{2} + {t}^{2} } = \dfrac{1}{a} {tan}^{ - 1} \bigg( \dfrac{t}{a}\bigg )}$

$\sf= > - \dfrac{1}{b} \dfrac{1}{a} {tan}^{ - 1} \bigg(\dfrac{t}{a}\bigg ) + c$

$\sf= > - \dfrac{1}{ab} {tan}^{ - 1} \bigg( \dfrac{bcosx}{a} \bigg) + c$

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Calculation of previous question given in attachment:

https://brainly.in/question/33322629?utm_source=android&utm_medium=share&utm_campaign=question