#### Explain solution rd sharma class 12 chapter Indefinite Integrals exercise 18.2 question 48

\begin{aligned} &2 \cos x+4 \sin x+1\\ &\text { Hint: Using } \int \sin x d x \text { and } \int \cos x d x\\ &\text { Given: } f^{\prime}(x)=a \sin x+b \cos x ; f^{\prime}(0)=4 ; f(0)=3, f\left(\frac{\pi}{2}\right)=5\\ &\text { Solution: } f^{\prime}(x)=a \sin x+b \cos x\\ &\text { Integrating both sides. }\\ &f(x)=\int a \sin x d x+\int b \cos x d x\\ &f(x)=a[-\cos x]+b[\sin x]+c\\ &f(x)=-a \cos x+b \sin x+c \quad \ldots \ldots .(*) \end{aligned}

\begin{aligned} &\mathrm{f}(0)=-a \cos 0+b \sin 0+c=3 \\ &\Rightarrow-a(1)+b(0)+c=3 \\ &\Rightarrow-a+c=3 \quad \ldots \ldots(1) \\ &f\left(\frac{\pi}{2}\right)=-a \cos \frac{\pi}{2}+b \sin \frac{\pi}{2}+c=5 \\ &=-a(0)+b(1)+c=5 \Rightarrow b+c=5 \end{aligned}

\begin{aligned} &\text { Also } f^{\prime}(0)=4 \\ &\text { i.e } a \sin 0+b \cos 0=4 \Rightarrow a(0)+b(1)=4 \Rightarrow b=4\\ &Put \;\;in (2)\\ &\text { i.e } b+c=5 \\ &=4+c=5 \Rightarrow c=1\\ &Put\;\; in(1)\\ &\text { i.e }-a+c=3 \\ &\Rightarrow-a+1=3 \Rightarrow-a=2 \Rightarrow a=-2\\ & \therefore \;\; By\;\; putting \;\;all\;\; the \;\;values\;\; in(*) ,\;\; we \;\;get\\ &f(x)=-(-2) \cos x+4 \sin x+c \\ &\Rightarrow f(x)=2 \cos x+4 \sin x+1 \end{aligned}