$\int 4 \cos \left(x + \frac{\pi}{6}\right) \cos 2x . \cos\left(\frac{5\pi}{6} + x\right)dx $
- $-\left(x - \frac{\sin4x}{4} - \frac{\sin2x}{2}\right) +c $
- $-\left(x + \frac{\sin4x}{4} - \frac{\sin2x}{2}\right) +c $
- $-\left(x - \frac{\sin4x}{4} + \frac{\sin2x}{2}\right) +c $
- $-\left(x + \frac{\sin4x}{4} + \frac{\sin2x}{2}\right) +c $
The Correct Option is A
Solution and Explanation
$=2 \int\left(\cos (2 x+\pi) \cos \frac{2 \pi}{3}\right) \cos\, 2 x\, d x$
$=2 \int\left(-\cos 2 x-\frac{1}{2}\right) \cos\, 2 x\, d x$
$=\int\left(-2 \cos ^{2} 2 x-\cos 2 x\right) d x$
$=-\int(1+\cos 4 x+\cos 2 x) d x$
$=-x-\frac{\sin 4 x}{x}-\frac{\sin 2 x}{2}+c$
Top Questions on Methods of Integration
Find the area of the region (in square units) enclosed by the curves: \[ y^2 = 8(x+2), \quad y^2 = 4(1-x) \] and the Y-axis.
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Evaluate the integral: \[ I = \int_{\frac{1}{\sqrt[5]{32}}}^{\frac{1}{\sqrt[5]{31}}} \frac{1}{\sqrt[5]{x^{30} + x^{25}}} dx. \]
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Questions Asked in BITSAT exam
- Let \( ABC \) be a triangle and \( \vec{a}, \vec{b}, \vec{c} \) be the position vectors of \( A, B, C \) respectively. Let \( D \) divide \( BC \) in the ratio \( 3:1 \) internally and \( E \) divide \( AD \) in the ratio \( 4:1 \) internally. Let \( BE \) meet \( AC \) in \( F \). If \( E \) divides \( BF \) in the ratio \( 3:2 \) internally then the position vector of \( F \) is:
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- Let \( \mathbf{a} = \hat{i} - \hat{k}, \mathbf{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}, \mathbf{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k} \). Then, \( [\mathbf{a} \, \mathbf{b} \, \mathbf{c}] \) depends on:}
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- Let the foot of perpendicular from a point \( P(1,2,-1) \) to the straight line \( L : \frac{x}{1} = \frac{y}{0} = \frac{z}{-1} \) be \( N \). Let a line be drawn from \( P \) parallel to the plane \( x + y + 2z = 0 \) which meets \( L \) at point \( Q \). If \( \alpha \) is the acute angle between the lines \( PN \) and \( PQ \), then \( \cos \alpha \) is equal to:
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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