Question 19 of 246MediumMCQ
The value of ∫−π/6π/6(1−sin(∣x∣+π/6)π+4x11)dx is equal to: Solutions · 1
Official solution
∫6π01−sin(∣x∣+6π)π+4x11+1−sin(∣x∣+6π)π−4x11dx=2π∫6π01−sin(x+6π)1dx=2π∫6π0cos2(x+6π)1+sin(x+6π)dx=2π∫6π0sec2(x+6π)+tan(x+6π)sec(x+6π)dx⇒2π[tan(x+6π)+sec(x+6π)]6π0⇒4π Answer the question to unlock