| 1. |
if\(\displaystyle\int\dfrac{\sin x}{\cos (x-a)}dx=Ax+B\log |sec(x-a)|+c\)where A, B and c are real constants then:1. A = sin a, B∈ℝ, c = cos a2. A = cos a, B = sin a, c∈ℝ3. A∈ℝ, B = cos a, c = sin a4. A = sin a, B = cos a, c∈ℝ |
|
Answer» Correct Answer - Option 4 : A = sin a, B = cos a, c∈ℝ Calculation: Given: \(\displaystyle\int\dfrac{\sin x}{\cos (x-a)}dx=Ax+B\log |sec(x-a)|+c\) \(Let = \int\frac{sin\ x}{sin (x - a)}dx\) Put t = x - a Differentiating w. r. t. x ⇒\(\frac{dt}{dx}=\frac{d(x-a)}{dx}\) ⇒\(\frac{dt}{dx}=1\) ⇒\(dx = dt\) ⇒\(\int\frac{sin(t + a)}{sin\ t}dt\) ⇒\(\int\frac{sin\ t\ cos\ a \ +\ cos\ t \ sin\ a}{sin\ t}dt \) \([sin (A+ B)= sin A cosB + cosA sinB]\) ⇒\(\int\frac{sint \ cosa }{sint}+\frac{cost\ sina}{sint}dt\) ⇒\(\int{cos\ a \ dt }+\int{cot \ t\ sina\ dt}\) ⇒\(cos a \int{dt + sin a}+ \int {cot \ t\ dt}\) (since\(sin a, cosa\)are constant) ⇒\(cos a ×t + sin a\ log |sint |+c\) ⇒ putting back t = x - a ⇒\({(x - a)}cos a+sina[sin(x-a)]+ c\) ⇒\(sin a\ log |sin (x-a)+ xcos a- a cosa + c\) ⇒\(sin a\ log|sin(x-a)|+xcosa + c_1\) \((c_1 = - acosa + c)\) |
|