Integrate the function: \(\frac{2\cos x-3\sin x}{6 \cos x+ 4\sin x}\)
Solution and Explanation
\(\frac{2\cos x-3\sin x}{6 \cos x+ 4\sin x}=\frac{2\cos x-3\sin x}{2(3\cos x+2\sin x)}\)
Let 3cos x+ 2sin x=t
∴ (-3sin x + 2cos x)dx = dt
\(\int \frac{2\cos x-3\sin x}{6 \cos x+ 4\sin x}dx=\int\frac{dt}{2t}\)
\(\frac{1}{2}\int\frac{1}{t}dt\)
= \(\frac{1}{2}\)log|t|+C
= \(\frac{1}{2}\)|2sin x + 3cos x|+C
Top Questions on integral
- Let \( y = f(x) \) be a thrice differentiable function in \( (-5, 5) \). Let the tangents to the curve \( y = f(x) \) at \( (1, f(1)) \) and \( (3, f(3)) \) make angles \( \frac{\pi}{6} \) and \( \frac{\pi}{4} \), respectively, with the positive x-axis. If \(2 \int_{\frac{1}{\sqrt{3}}}^{1} \left( \left( f'(t) \right)^2 + 1 \right) f''(t) \, dt = \alpha + \beta \sqrt{3}\) where \( \alpha \), \( \beta \) are integers, then the value of \( \alpha + \beta \) equals
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Questions Asked in CBSE CLASS XII exam
- Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)
- For what values of x,\(\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\)\(\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 & 1 \\1&0&2 \end{bmatrix}\)\(\begin{bmatrix} 0 \\2\\x\end{bmatrix}\)=O?
- Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\)
What is the Planning Process?
- CBSE CLASS XII - 2023
- Planning process steps
- Find the inverse of each of the matrices,if it exists. \(\begin{bmatrix} 2 & 3\\ 5 & 7 \end{bmatrix}\)
Concepts Used:
Methods of Integration
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
Integration by Parts:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
Method of Integration Using Partial Fractions:
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Integration by Substitution Method
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
Integration by Decomposition
Reverse Chain Rule
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C