JEE Main 2024 · 5 April · morning · Single correct · +4 / −1
The integral ∫0^(π / 4) (136 sin x)/(3 sin x+5 cos x) d x is equal to :
- (a) 3 π-50 log ₑ 2+20 log ₑ 5
- (b) 3 π-25 log ₑ 2+10 log ₑ 5
- (c) 3 π-10 log ₑ(2 √2)+10 log ₑ 5
- (d) 3 π-30 log ₑ 2+20 log ₑ 5
Answer and solution
Answer (a) 3 π-50 log ₑ 2+20 log ₑ 5
Solution
Maths sir · 4 steps
Write the numerator as A times the denominator plus B times its derivative, and it splits.
- Split the numerator. Let D = 3 sin x + 5 cos x, so D' = 3 cos x - 5 sin x. Write sin x = A D + B D' and match coefficients: 3A - 5B = 1 from sin x and 5A + 3B = 0 from cos x. This gives A = 3/34 and B = -5/34.
(136 sin x)/(3 sin x + 5 cos x) = 12 - 20 (3 cos x - 5 sin x)/(3 sin x + 5 cos x)
- Integrate each part. The constant gives 12 × π/4 = 3π. The second part is of the form D'/D, so it integrates to -20[ln(3 sin x + 5 cos x)]₀^(π/4).
- Put in limits. At x = π/4, D = 3/√2 + 5/√2 = 8/√2 = 4√2. At x = 0, D = 5. So the log part is -20 ln(4√2) + 20 ln 5, and ln(4√2) = 5/2 ln 2.
- So. The integral is 3π - 50 logₑ 2 + 20 logₑ 5, which is option (a).
The idea to keep. For ∫ (p sin x + q cos x)/(a sin x + b cos x) dx, always write the numerator as A times the denominator plus B times its derivative. You get Ax + B ln|denominator|.
Answer from the NTA final answer key. Solution by PrepWiser.