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Question: How do you evaluate \( \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx}\) ?...

How do you evaluate ∫cos⁡x9+sin⁡2xdx \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx} ?

Explanation

Solution

We should solve this problem by using substitutions. Here we can substitute sin x as y. Then you can use some formulas to solve the integral and finally get the answer to this integral.

Complete step by step solution:
According to the problem, we are asked to find the integral of the given equation, that is ∫cos⁡x9+sin⁡2xdx \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx}. We take the equation as equation 1.

∫cos⁡x9+sin⁡2xdx \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx}--- ( 1 )

Therefore, by using the substitution, where we substitute sinx with y, we get
x=sin⁡xx=\sin x ⇒sin⁡−1x=y \Rightarrow {{\sin }^{-1}}x=y ---- (2)

∴dx=cos⁡ydy \therefore dx=\cos ydy---- (3)

Using the equation 2 and equation 3, if we substitute them in equation 1, we get:
⇒∫cos⁡x9+sin⁡2xdx=∫19+y2dy  \begin{aligned} & \Rightarrow \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx}=\int{\dfrac{1}{9+{{y}^{2}}}dy} \\\ & \\\ \end{aligned}

Here, we can now divide with 1/9 on both the numerator and the denominator in the above equation.
⇒∫cos⁡x9+sin⁡2xdx=19∫11+y29dy\Rightarrow \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx}=\dfrac{1}{9}\int{\dfrac{1}{1+\dfrac{{{y}^{2}}}{9}}dy}

⇒∫cos⁡x9+sin⁡2xdx=19∫11+(y3)2dy \Rightarrow \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx}=\dfrac{1}{9}\int{\dfrac{1}{1+{{\left( \dfrac{y}{3} \right)}^{2}}}dy} --- (3)

Now, we again use substitution to substitute y3 \dfrac{y}{3} with z. Therefore, we get:

⇒y3=z⇒dy=3dz \Rightarrow \dfrac{y}{3}=z\Rightarrow dy=3dz ---- (4)

Now, after substituting the value in 4 in equation 3, we get:
⇒∫cos⁡x9+sin⁡2xdx=19∫31+z2dz\Rightarrow \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx}=\dfrac{1}{9}\int{\dfrac{3}{1+{{z}^{2}}}dz}

⇒∫cos⁡x9+sin⁡2xdx=13∫11+z2dz \Rightarrow \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx}=\dfrac{1}{3}\int{\dfrac{1}{1+{{z}^{2}}}dz} ---- (5)

We know that d(tan⁡−1x)dx=11+x2   \begin{aligned} & \dfrac{d\left( {{\tan }^{-1}}x \right)}{dx}=\dfrac{1}{1+{{x}^{2}}} \\\ & \\\ \end{aligned}. Therefore, using this in equation 5, we get:

⇒∫cos⁡x9+sin⁡2xdx=13tan⁡−1(z)+c=13tan⁡−1(y3)+c=13tan⁡−1(sin⁡x3)+c \Rightarrow \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx}=\dfrac{1}{3}{{\tan }^{-1}}\left( z \right)+c=\dfrac{1}{3}{{\tan }^{-1}}\left( \dfrac{y}{3} \right)+c=\dfrac{1}{3}{{\tan }^{-1}}\left( \dfrac{\sin x}{3} \right)+c
⇒∫cos⁡x9+sin⁡2xdx=13tan⁡−1(sin⁡x3)+c\Rightarrow \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx}=\dfrac{1}{3}{{\tan }^{-1}}\left( \dfrac{\sin x}{3} \right)+c ---- Final answer

So, we have found the derivative of the given equation ∫cos⁡x9+sin⁡2xdx \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx} as ∫cos⁡x9+sin⁡2xdx=13tan⁡−1(sin⁡x3)+c \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx}=\dfrac{1}{3}{{\tan }^{-1}}\left( \dfrac{\sin x}{3} \right)+c.

Therefore, the solution of the given equation ∫cos⁡x9+sin⁡2xdx \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx} is ∫cos⁡x9+sin⁡2xdx=13tan⁡−1(sin⁡x3)+c \int{\dfrac{\cos x}{9+{{\sin }^{2}}x}dx}=\dfrac{1}{3}{{\tan }^{-1}}\left( \dfrac{\sin x}{3} \right)+c.

Note: We should be careful while doing all the substitutions. If one substitution goes wrong, the total answer may go wrong. Also, if you want to verify your answer, you could differentiate the answer you got. If after differentiating you get back the question as your answer, then your answer is correct otherwise check where you went wrong.