Animated Solution for Mathematics - Indefinite Integration: ∫cosx+3sinxdx equals
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Visualized Solution
Identifying the Form acosx+bsinx
Given integral: ∫cosx+3sinxdx
The denominator is of the form acosx+bsinx
Our goal is to simplify this linear combination into a single trigonometric term.
Identifying Coefficients a and b
Comparing cosx+3sinx with acosx+bsinx:
Here, the coefficient of cosx is a=1
The coefficient of sinx is b=3
The Right Triangle Analogy
We can represent these coefficients geometrically.
Let the base of a right-angled triangle be b=3
Let the perpendicular height be a=1
Calculating the Hypotenuse R
The scaling factor R is the hypotenuse of this triangle:
R=a2+b2
R=12+(3)2=1+3=4=2
Finding the Phase Angle θ
Let θ be the angle opposite to the height a=1:
sinθ=HypotenusePerpendicular=21
cosθ=HypotenuseBase=23
This gives θ=6π (or 30∘)
Transforming the Denominator
Multiply and divide the denominator by R=2:
cosx+3sinx=2[21cosx+23sinx]
Applying Trigonometric Identity
Substitute 21=sin(6π) and 23=cos(6π):
Denominator =2[sin(6π)cosx+cos(6π)sinx]
Using sin(A+B)=sinAcosB+cosAsinB:
Denominator =2sin(x+6π)
Substituting Back into the Integral
Substitute the simplified denominator back into the integral:
I=∫2sin(x+6π)dx
Factor out the constant and use sinθ1=cscθ:
I=21∫csc(x+6π)dx
Applying the Cosecant Integration Formula
Standard formula: ∫cscudu=logtan(2u)+C
Here, u=x+6π
Therefore, 2u=2x+12π
Final Simplification and Result
Applying the formula gives:
I=21logtan(2x+12π)+C
This matches Option 3.
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The Sigma Insight: Integration by Substitution
Solution Diagram
The Art of Trigonometric Transformation
Mastering the Linear Combination
Welcome, fellow traveler on the path to JEE mastery. Today, we are going to demystify one of the most elegant techniques in calculus: the transformation of a linear combination of sine and cosine.
When you first look at the integral I=∫cosx+3sinxdx, it might seem like a standard, perhaps even boring, problem. But beneath the surface lies a beautiful geometric reality that, once understood, will change how you approach trigonometric integrals forever.
Phase 1
The Geometric Intuition
Whenever you encounter an expression of the form acosx+bsinx, do not panic. Instead, think of it as a vector.
Imagine a right-angled triangle where the base is b=3 and the height is a=1. The hypotenuse of this triangle, which we call R, is given by the Pythagorean theorem:
R=a2+b2=12+(3)2=1+3=2
This R is our scaling factor. By multiplying and dividing our denominator by 2, we force the expression into a form that mirrors the compound angle identity sin(A+B)=sinAcosB+cosAsinB.
Phase 2
The Transformation
We rewrite the denominator as:
cosx+3sinx=2[21cosx+23sinx]
Now, look closely at the terms inside the bracket. We know that sin(6π)=21 and cos(6π)=23.
Substituting these values, the expression becomes:
2[sin(6π)cosx+cos(6π)sinx]
This is the exact expansion of sin(x+6π). Our denominator has been tamed; it is now simply 2sin(x+6π).
Phase 3
The Final Integration
With the denominator simplified, our integral becomes:
I=∫2sin(x+6π)dx=21∫csc(x+6π)dx
I know that seeing a cosecant integral might make you nervous, but remember the standard formula: ∫cscudu=log∣tan(2u)∣+C. Here, our u is x+6π, and therefore 2u becomes 2x+12π.
Applying the formula, we arrive at our final, elegant result:
I=21logtan(2x+12π)+C
Conclusion
The Beauty of the Method
This problem is a perfect example of why we study mathematics. It is not just about memorizing formulas; it is about recognizing patterns and using geometry to simplify the complex.
By transforming a sum of two functions into a single function, we turned a potentially messy integration into a simple, standard form. Keep this geometric triangle analogy in your toolkit—it will serve you well in many more JEE problems to come.
You have the power to break down any problem; just take a breath, visualize the geometry, and let the math flow.