Integrate:Sinxcosx/a^2cos^2(x)+b^2sin^2(x))dx???
plz i need the full working too hard for me
∫(sinx cosx)/(a^2 cos^2(x) + b^2 sin^2(x)) dx
well, let's see.
a^2 cos^2(x) + b^2 sin^2(x)
= a^2 cos^2(x) + a^2 sin^2(x) + (b^2-a^2)sin^2(x)
= a^2 + (b^2-a^2) sin^2(x)
so, if
u = a^2 + (b^2-a^2) sin^2(x)
du = 2(b^2-a^2) sinx cosx dx
and your integral now becomes
1/2 ∫ du/u
Not so hard now, eh?
Or, I guess you could just let
u = a^2 cos^2(x) + b^2 sin^2(x)
du = -2a^2 sinx cosx + 2b^2 sinx cosx
= 2sinx cosx (b^2-a^2)
and you have
1/(2(b^2-a^2)) ∫ du/u
To solve the given integral \(\int \frac{\sin(x)\cos(x)}{a^2\cos^2(x)+b^2\sin^2(x)}dx\), we can follow these steps:
Step 1: Simplify the expression in the denominator
Using the trigonometric identity \(\sin^2(x) + \cos^2(x) = 1\), we can rewrite the denominator as:
\(a^2\cos^2(x) + b^2\sin^2(x) = a^2(1 - \sin^2(x)) + b^2\sin^2(x)\)
\(= a^2 - a^2\sin^2(x) + b^2\sin^2(x)\)
\(= a^2 + (b^2 - a^2)\sin^2(x)\)
So, our integral becomes: \(\int \frac{\sin(x)\cos(x)}{a^2 + (b^2 - a^2)\sin^2(x)}dx\)
Step 2: Apply a substitution
Let \(u = \sin(x)\). Then \(du = \cos(x)dx\).
Step 3: Substitute and simplify the integral
Using the substitution \(u = \sin(x)\) and \(du = \cos(x)dx\), we convert our integral to:
\(\int \frac{du}{a^2 + (b^2 - a^2)u^2}\)
Step 4: Apply partial fraction decomposition
To simplify the integrand further, we can decompose it into partial fractions. The denominator can be factored as \((a^2 + (b^2 - a^2)u^2) = (a^2 + (b^2 - a^2)u)(a^2 + (b^2 - a^2)(-u))\).
Thus, the integral becomes a sum of simpler integrals:
\(\int \left(\frac{k_1}{a^2 + (b^2 - a^2)u} + \frac{k_2}{a^2 + (b^2 - a^2)(-u)}\right) du\)
Step 5: Solve for the constants and simplify
By finding a common denominator and equating the numerators of the fractions, we can obtain the values of \(k_1\) and \(k_2\).
Step 6: Evaluate the integral
After finding the constants, we integrate each term separately. The integral of \(\frac{k_1}{a^2 + (b^2 - a^2)u}\) can be evaluated as \(\frac{k_1}{\sqrt{b^2 - a^2}} \arctan\left(\frac{\sqrt{b^2 - a^2}u}{a}\right)\), and the integral of \(\frac{k_2}{a^2 + (b^2 - a^2)(-u)}\) can be evaluated similarly.
Step 7: Substitute back the variable
Replace \(u\) with \(\sin(x)\) in the final result.
I hope this explanation helps you understand the process of solving the given integral.