Hi Maths Lover!

Given: An expression [tex]\int\limits^\frac{\pi }{2} _\frac{-\pi }{2} ({x^{3} + cosx+tan^{5}x }) \, dx[/tex]

To Find: The Simplified Value of this expression

Solution:

• We are given the expression [tex]\int\limits^\frac{\pi }{2} _\frac{-\pi }{2} ({x^{3} + cosx+tan^{5}x }) \, dx[/tex].

• Use the property of Integral [tex]\int\limits f(x) \pm g(x) dx = \int\limits f(x) dx \pm \int\limits g(x) dx[/tex] .

[tex]\int\limits^\frac{\pi }{2} _\frac{-\pi }{2} {x^{3} dx + \int\limits^\frac{\pi }{2} _\frac{-\pi }{2} cos(x) dx + \int\limits^\frac{\pi }{2} _\frac{-\pi }{2} {tan(x)^{5} dx[/tex]

• The function [tex]f(x) = x^{3}[/tex] is odd, so use the property of odd functions, [tex]\int\limits^a_{-a} f({x}) \, dx = 0[/tex].

[tex]0+ \int\limits^\frac{\pi }{2} _\frac{-\pi }{2} cos(x) dx + \int\limits^\frac{\pi }{2} _\frac{-\pi }{2} {tan(x)^{5} dx[/tex]

• The function [tex]f(x) = cos(x)[/tex] is even, so use the property of functions,

• [tex]\int\limits^a_{-a} f({x}) \, dx = 2(\int\limits^a_{-a} f({x}) \, dx )[/tex]

[tex]0+ 2 \times(\int\limits^\frac{\pi }{2} _0 cos(x) dx) + \int\limits^\frac{\pi }{2} _\frac{-\pi }{2} {tan(x)^{5} dx[/tex]

• The function [tex]f(x) = tan(x)^{5}[/tex] is odd, so use the property of odd functions, [tex]\int\limits^a_{-a} f({x}) \, dx = 0[/tex].

[tex]0+ 2 \times(\int\limits^\frac{\pi }{2} _0 cos(x) dx) + 0[/tex]

• Removing [tex]0[/tex] does not change the value, so remove it from the expression.

[tex]2 \times(\int\limits^\frac{\pi }{2} _0 cos(x) dx) + 0[/tex]

• Since removing [tex]0[/tex] does not change the value, remove the other [tex]0[/tex] also.

[tex]2 \times(\int\limits^\frac{\pi }{2} _0 cos(x) dx)[/tex]

• We know that the value of definite integral  [tex]\int\limits^\frac{\pi }{2} _0 cos(x) dx[/tex]  is [tex]1[/tex]. So, we can write:

[tex]2\times1[/tex]

• Multiply and Calculate:

[tex]=2[/tex]

Final Answer:

Hence, the value of the integral  [tex]\int\limits^\frac{\pi }{2} _\frac{-\pi }{2} ({x^{3} + cosx+tan^{5}x }) \, dx[/tex]  is  [tex]2[/tex].

Hope it helps!

Keep Learning!

Thanks!

Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given integral is

[tex]\sf \: \displaystyle \sf\int_{-\pi / 2}^{\pi / 2}\left(x^{3}+\cos x+\tan ^{5} x\right) d x \\ \\ [/tex]

can be rewritten as

[tex]\sf \: = \: \displaystyle \sf\int_{-\pi / 2}^{\pi / 2} {x}^{3}dx \: + \: \displaystyle \sf\int_{-\pi / 2}^{\pi / 2}cosx \: dx \: + \: \displaystyle \sf\int_{-\pi / 2}^{\pi / 2} {tan}^{5} x \: dx \\ \\ [/tex]

Now, Let assume that,

[tex]\sf \: f(x) = {x}^{3} \\ \\ [/tex]

[tex]\sf \: g(x) = cosx \\ \\ [/tex]

[tex]\sf \: h(x) = {tan}^{5}x \\ \\ [/tex]

Now, Consider

[tex]\bf \: f( - x) = {( - x)}^{3} = - {x}^{3} = - f(x) \\ \\ [/tex]

[tex]\bf \: g( - x) = cos( - x) = cosx = g(x) \\ \\ [/tex]

[tex]\bf \: h( - x) = {tan}^{5}( - x) = - {tan}^{5}x \: = \: - h(x) \\ \\ [/tex]

We know that,

[tex]\boxed{ \sf{ \:\begin{gathered}\begin{gathered}\bf\: \displaystyle \sf\int_{-a}^{a} f(x) \: dx= \begin{cases} &\sf{2\displaystyle\sf\int_{0}^{a} f(x) \: dx , \: \: \: if \: f( - x) = f(x)} \\ &\sf{0 ,\: \: if \: f( - x) = - f(x)} \end{cases}\end{gathered}\end{gathered} \: }} \\ \\ [/tex]

So, using these, we get

[tex]\sf \: = \: 0 \: + \: 2\displaystyle \sf\int_{0}^{\pi / 2}cosx \: dx \: + \: 0 \\ \\ [/tex]

[tex]\sf \: = \: 2\bigg(sinx \bigg) _{0}^{\pi / 2} \\ \\ [/tex]

[tex]\sf \: = \: 2\bigg(sin \frac{\pi}{2} - sin0 \bigg) \\ \\ [/tex]

[tex]\sf \: = \: 2\bigg(1 - 0 \bigg) \\ \\ [/tex]

[tex]\sf \: = \: 2 \\ \\ [/tex]

Hence,

[tex]\bf\implies \:\displaystyle \bf\int_{-\pi / 2}^{\pi / 2}\left(x^{3}+\cos x+\tan ^{5} x\right) d x = 2 \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

[tex]\boxed{ \rm{ \:\displaystyle \int_{a}^{b}\sf \: f(x) \: dx \: = - \: \: \displaystyle \int_{b}^{a}\sf \: f(x) \: dx}} \\ [/tex]

[tex]\boxed{ \rm{ \:\displaystyle \int_{a}^{b}\sf \: f(x) \: dx \: = \: \: \displaystyle \int_{a}^{b}\sf \: f(y) \: dy}} \\ [/tex]

[tex]\boxed{ \rm{ \:\displaystyle \int_{0}^{a}\sf \: f(x) \: dx \: = \: \: \displaystyle \int_{0}^{a}\sf \: f(a - x) \: dx}} \\ [/tex]

[tex]\boxed{ \rm{ \:\displaystyle \int_{a}^{b}\sf \: f(x) \: dx \: = \: \: \displaystyle \int_{a}^{b}\sf \: f(a + b - x) \: dx}} \\ [/tex]

[tex]\boxed{ \rm{ \:\displaystyle \int_{0}^{2a}\sf \: f(x) \: dx \: = \:2 \displaystyle \int_{0}^{2}\sf f(x)dx \: \: if \: f(2a - x) = f(x)}} \\ [/tex]

[tex]\boxed{ \rm{ \:\displaystyle \int_{0}^{2a}\sf \: f(x) \: dx \: = \:0 \: \: if \: f(2a - x) = - \: f(x)}} \\ [/tex]