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Question: How do you evaluate the integral \(\int {3\sin x + 4\cos xdx} \)?...

How do you evaluate the integral ∫3sin⁡x+4cos⁡xdx\int {3\sin x + 4\cos xdx} ?

Explanation

Solution

First, use the property that the integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the various functions. Next, use the property that the integral of the product of a constant and a function = the constant ×\times integral of the function. Then, use integration formulas (III) to get the required result.

Formula used:
The integral of the product of a constant and a function = the constant ×\times integral of the function.
i.e., ∫(kf(x)dx)=k∫f(x)dx\int {\left( {kf\left( x \right)dx} \right)} = k\int {f\left( x \right)dx} , where kk is a constant.
The integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the various functions.
i.e., ∫[f(x)±g(x)]dx=∫f(x)dx±∫g(x)dx\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int {f\left( x \right)dx} \pm \int {g\left( x \right)dx}
Integration formula: ∫sin⁡xdx=−cos⁡x+C\int {\sin xdx} = - \cos x + C and ∫cos⁡xdx=sin⁡x+C\int {\cos xdx} = \sin x + C

Complete step by step solution:
We have to find ∫3sin⁡x+4cos⁡xdx\int {3\sin x + 4\cos xdx} …(i)
First, using the property that the integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the various functions.
i.e., ∫[f(x)±g(x)]dx=∫f(x)dx±∫g(x)dx\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int {f\left( x \right)dx} \pm \int {g\left( x \right)dx}
So, in integral (i), we can use above property
∫3sin⁡x+4cos⁡xdx=∫3sin⁡dx+∫4cos⁡xdx\int {3\sin x + 4\cos xdx} = \int {3\sin dx} + \int {4\cos xdx} …(ii)
Now, using the property that the integral of the product of a constant and a function = the constant ×\times integral of the function.
i.e., ∫(kf(x)dx)=k∫f(x)dx\int {\left( {kf\left( x \right)dx} \right)} = k\int {f\left( x \right)dx} , where kk is a constant.
So, in integral (ii), we can use above property
∫3sin⁡x+4cos⁡xdx=3∫sin⁡dx+4∫cos⁡xdx\int {3\sin x + 4\cos xdx} = 3\int {\sin dx} + 4\int {\cos xdx} …(iii)
Now, using the integration formula ∫sin⁡xdx=−cos⁡x+C\int {\sin xdx} = - \cos x + C and ∫cos⁡xdx=sin⁡x+C\int {\cos xdx} = \sin x + C in integral (iii), we get
∫3sin⁡x+4cos⁡xdx=3(−cos⁡x+C1)+4(sin⁡x+C2)\int {3\sin x + 4\cos xdx} = 3\left( { - \cos x + {C_1}} \right) + 4\left( {\sin x + {C_2}} \right)
Final solution: Hence, ∫3sin⁡x+4cos⁡xdx=−3cos⁡x+4sin⁡x+C\int {3\sin x + 4\cos xdx} = - 3\cos x + 4\sin x + C.

Note: Here, we combined the two constants of integration, C1{C_1} and C2{C_2} into one constant CC because it’s easier to deal with.
In Maths, integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts. This method is used to find the summation under a vast scale. Calculation of small addition problems is an easy task which we can do manually or by using calculators as well. But for big additional problems, where the limits could reach to even infinity, integration methods are used. Integration and differentiation both are important parts of calculus. The concept level of these topics is very high.