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

\({(x - a)}cos a+sina[sin(x-a)]+ c\)

\(sin a\ log |sin (x-a)+ xcos a- a cosa + c\)

\(sin a\ log|sin(x-a)|+xcosa + c_1\) \((c_1 = - acosa + c)\)



Discussion

No Comment Found

Related InterviewSolutions